Series: Determine if they are convergent or divergent

[jlmccart03]

Homework Statement

I have a couple of series where I need to find out if they are convergent (absolute/conditional) or divergent.

• Σ(n³/3ⁿ
  
• Σk(2/3)^k
  
• Σ√n/1+n²
  
• Σ(-1)ⁿ⁺¹*n/n^2+9

Homework Equations

Comparison Test
Ratio Test
Alternating Series Test
Divergence Test, etc

The Attempt at a Solution

For the first series I determined it converges absolutley since taking the ratio test gives me 1/3 which is less than 1.
The second series I am lost on, I tried comparison test for a geometric series, but that k is an issue, so I went to ratio test, which is even more confusing for this series. I want to know what series test I should use and then I will do the work.
The third series I just compared 1/n² and got taht it converges since we know the Σ1/n² converges and the series a_n < b_n (where a_n is the third series in the bullet point list and b_n is my comparison series.
The final series I did the same thing, but since it is alternating I decided to go with the alternating series test which I really don't know how to solve, but overall the series diverged if I did the test for divergence where lim n→∞ (n/n²+9) ≠ 0.

Am I correct on most of these and could I get help with series 2.

 

[Mark44]

Homework Statement

I have a couple of series where I need to find out if they are convergent (absolute/conditional) or divergent.

• Σ(n³/3ⁿ
  
• Σk(2/3)^k
  
• Σ√n/1+n²
  
• Σ(-1)ⁿ⁺¹*n/n^2+9

Be more careful with parentheses. In the first series above, you're missing a right paren. In the third, I assume that 1 + n² is the denominator, so the series should be written as Σ√n/(1+n²). The last series should probably be written as Σ(-1)ⁿ⁺¹*n/(n^2+9)

Homework Equations

Comparison Test
Ratio Test
Alternating Series Test
Divergence Test, etc

The Attempt at a Solution

For the first series I determined it converges absolutley since taking the ratio test gives me 1/3 which is less than 1.
The second series I am lost on, I tried comparison test for a geometric series, but that k is an issue, so I went to ratio test, which is even more confusing for this series. I want to know what series test I should use and then I will do the work.

The Ratio Test will work. Please show us what you did.

The third series I just compared 1/n² and got taht it converges since we know the Σ1/n² converges and the series a_n < b_n (where a_n is the third series in the bullet point list and b_n is my comparison series.

If that worked for you, and you were able to show that the terms in the third series are all smaller than the corresponding terms in $\sum \frac 1 {n^2}$, then fine. Personally, I would have chosen $\sum \frac 1{n^{3/2}}$ for the comparison, or would have used the limit comparison test using the same series.

The final series I did the same thing, but since it is alternating I decided to go with the alternating series test which I really don't know how to solve, but overall the series diverged if I did the test for divergence where lim n→∞ (n/n²+9) ≠ 0.

But as you point out, the fourth one is an alternating series. The alternating series test is fairly easy to use. What does it say on the fourth series?

Am I correct on most of these and could I get help with series 2.

 

[Ray Vickson]

Homework Statement

I have a couple of series where I need to find out if they are convergent (absolute/conditional) or divergent.

• Σ(n³/3ⁿ
  
• Σk(2/3)^k
  
• Σ√n/1+n²
  
• Σ(-1)ⁿ⁺¹*n/n^2+9

Homework Equations

Comparison Test
Ratio Test
Alternating Series Test
Divergence Test, etc

The Attempt at a Solution

For the first series I determined it converges absolutley since taking the ratio test gives me 1/3 which is less than 1.
The second series I am lost on, I tried comparison test for a geometric series, but that k is an issue, so I went to ratio test, which is even more confusing for this series. I want to know what series test I should use and then I will do the work.
The third series I just compared 1/n² and got taht it converges since we know the Σ1/n² converges and the series a_n < b_n (where a_n is the third series in the bullet point list and b_n is my comparison series.
The final series I did the same thing, but since it is alternating I decided to go with the alternating series test which I really don't know how to solve, but overall the series diverged if I did the test for divergence where lim n→∞ (n/n²+9) ≠ 0.

Am I correct on most of these and could I get help with series 2.

The ratio test works just fine on series 2.

For series 3: what you wrote means
$$\sum \frac{\sqrt{n}}{1} + n^2 + 1,$$
which is obviously divergent. However, maybe you did not really mean what you wrote; perhaps you meant
$$\sum \frac{\sqrt{n}}{n^2+1}\hspace{1cm}(1)$$
or maybe you meant
$$ \sum \sqrt{\frac{n}{n^2+1}} \hspace{1cm}(2)$$
It is impossible to tell from what you wrote. You need to use parentheses.

Same type of problem with series 4: you wrote
$$\sum (-1)^{n+1} \frac{n}{n^2} + 1,$$
but maybe you meant
$$\sum (-1)^{n+1} \frac{n}{n^2+1}.$$
Again, use parentheses.

 

[jlmccart03]

Be more careful with parentheses. In the first series above, you're missing a right paren. In the third, I assume that 1 + n² is the denominator, so the series should be written as Σ√n/(1+n²). The last series should probably be written as Σ(-1)ⁿ⁺¹*n/(n^2+9)
The Ratio Test will work. Please show us what you did.
If that worked for you, and you were able to show that the terms in the third series are all smaller than the corresponding terms in $\sum \frac 1 {n^2}$, then fine. Personally, I would have chosen $\sum \frac 1{n^{3/2}}$ for the comparison, or would have used the limit comparison test using the same series.
But as you point out, the fourth one is an alternating series. The alternating series test is fairly easy to use. What does it say on the fourth series?

Ok, I will show the work after class. Is it ok if I post pictures of written work. I prefer it over electronic.

 

[Mark44]

Ok, I will show the work after class. Is it ok if I post pictures of written work. I prefer it over electronic.

We much prefer that you post your work as text, preferably using LaTeX (tutorial here). You can see some examples in Ray's post. Right-click on any of his examples, and then select Show Math As -- TeX Commands

 

[jlmccart03]

We much prefer that you post your work as text, preferably using LaTeX (tutorial here). You can see some examples in Ray's post. Right-click on any of his examples, and then select Show Math As -- TeX Commands

Why may I ask? I for one don't have time for learning the commands due to class and such, but more so I don't learn at all from electronic. It's not beneficial to me as I learn from doing physically, typing doesn't do that. I'm just curious as to the reason since this is a website meant to help you learn and I don't learn from electronic writing.

 

[Mark44]

Why may I ask? I for one don't have time for learning the commands due to class and such, but more so I don't learn at all from electronic. It's not beneficial to me as I learn from doing physically, typing doesn't do that. I'm just curious as to the reason since this is a website meant to help you learn and I don't learn from electronic writing.

Several reasons.
1) Students often post images that are sideways, or upside-down, or unreadable due to illegible handwriting or poor lighting when they took the picture. This happens a lot. Some helpers won't even bother replying in such cases.
2) If the work shown in an image has a mistake several lines down, we as helpers have to identify where the error is instead of being able to insert a comment directly at the place where the error is. You can't insert a comment in the middle of an image.

As you said, we intend to help students learn, but if you aren't willing to put in a very small amount of time to make it easier for us to help you, can't you see why some of us would likewise not be willing to put in extra effort to help you? Take a look at Ray's post, #3. He put in a considerable effort to make what you wrote very clear. In total, he used four commands in TeX: summation, square root, fraction, and exponent. All of these are covered, with examples, in the tutorial I linked to. I would guess that in 5 to 10 minutes, you could make a good start at learning some TeX.

You came here for help. Do us a courtesy by making it easier to help you, not harder.

 

[jlmccart03]

Several reasons.
1) Students often post images that are sideways, or upside-down, or unreadable due to illegible handwriting or poor lighting when they took the picture. This happens a lot. Some helpers won't even bother replying in such cases.
2) If the work shown in an image has a mistake several lines down, we as helpers have to identify where the error is instead of being able to insert a comment directly at the place where the error is. You can't insert a comment in the middle of an image.

As you said, we intend to help students learn, but if you aren't willing to put in a very small amount of time to make it easier for us to help you, can't you see why some of us would likewise not be willing to put in extra effort to help you? Take a look at Ray's post, #3. He put in a considerable effort to make what you wrote very clear. In total, he used four commands in TeX: summation, square root, fraction, and exponent. All of these are covered, with examples, in the tutorial I linked to. I would guess that in 5 to 10 minutes, you could make a good start at learning some TeX.

You came here for help. Do us a courtesy by making it easier to help you, not harder.

Alright, well I thank you for the effort you put into helping me, but I simply just don't have the time to learn these commands, I understand that some people may write horribly or post poor pictures, but I simply cannot learn through typing my work and answering questions through typed work. I understand that you guys don't have all the time to decipher what others may write in hand, but then again if this were a help room at a university then that is what you would be dealing with. Sorry to inconvienance you, but I cannot and will not type all of my work when I do not learn that way. I am here to learn mathematics and not write a bunch of commands that I simply don't have time to do at this moment. I came here to learn how to do the series, but since I have to apparently take time out of my learning to learn something that is completely off topic to what I asked I will simply search elsewhere.

 

[Ray Vickson]

Ok, I will show the work after class. Is it ok if I post pictures of written work. I prefer it over electronic.

We really discourage that; most helpers will not look at pictures of work, but you might get lucky and encounter one helper who is willing to overlook the issue.

Ok, I will show the work after class. Is it ok if I post pictures of written work. I prefer it over electronic.

Please do not post images of written work. You may prefer it, but most helpers will not even look at it if you do that.

Alright, well I thank you for the effort you put into helping me, but I simply just don't have the time to learn these commands, I understand that some people may write horribly or post poor pictures, but I simply cannot learn through typing my work and answering questions through typed work. I understand that you guys don't have all the time to decipher what others may write in hand, but then again if this were a help room at a university then that is what you would be dealing with. Sorry to inconvienance you, but I cannot and will not type all of my work when I do not learn that way. I am here to learn mathematics and not write a bunch of commands that I simply don't have time to do at this moment. I came here to learn how to do the series, but since I have to apparently take time out of my learning to learn something that is completely off topic to what I asked I will simply search elsewhere.

You don't need to learn new commands; just using parentheses is enough---not as good as using LaTeX, but acceptable.

The point is that some expressions are either ambiguous or just plain wrong without brackets. The expression a/b+c means $\frac{a}{b} + c$ when parsed using standard rules for reading math. So, if you want $\frac{a}{b+c}$ you need to enforce it, like this: a/(b+c). Easy: one extra keystroke! If you use the √ symbol, you definitely need parentheses. The expression √a/b means $\sqrt{a} \, /b= \frac{\sqrt{a}}{b},$ so if you want to write $\sqrt{\frac{a}{b}} = \sqrt{a/b}$ in plain text you need to enforce operation prioities by using parentheses, like this: √(a/b). Unfortunately, that can get you involved with multiple parentheses, as in √(a/(b+c)). In LaTeX that would be $\sqrt{a/(b+c)}$ with no need for extra brackets, because the horizontal line on the top of the square-root sign expands to cover the whole argument, such as in $\sqrt{2}$ vs. $\sqrt{22222}$. But still, you can write most of what you need in plain text; it just needs a tiny bit of extra work on your part.

 

[Mark44]

Alright, well I thank you for the effort you put into helping me, but I simply just don't have the time to learn these commands.

You don't have to use LaTeX. What you posted earlier is fine (provided that you include parentheses where they are needed). It's just that we discourage posting images of work, for the reasons I already mentioned.

I came here to learn how to do the series, but since I have to apparently take time out of my learning to learn something that is completely off topic to what I asked I will simply search elsewhere.

That's your call, of course. Keep in mind that we saved you a fair amount of time already in the questions you posted.

 

[jlmccart03]

We really discourage that; most helpers will not look at pictures of work, but you might get luPlease do not post images of written work. You may prefer it, but most helpers will not even look at it if you do that. (SYou don't need to learn new commands; just using parentheses is enough---not as good as using LaTeX, but acceptable.

The point is that some expressions are either ambiguous or just plain wrong without brackets. The expression a/b+c means $\frac{a}{b} + c$ when parsed using standard rules for reading math. So, if you want $\frac{a}{b+c}$ you need to enforce it, like this: a/(b+c). Easy: one extra keystroke! If you use the √ symbol, you definitely need parentheses. The expression √a/b means $\sqrt{a} \, /b= \frac{\sqrt{a}}{b},$ so if you want to write $\sqrt{\frac{a}{b}} = \sqrt{a/b}$ in plain text you need to enforce operation prioities by using parentheses, like this: √(a/b). Unfortunately, that can get you involved with multiple parentheses, as in √(a/(b+c)). In LaTeX that would be $\sqrt{a/(b+c)}$ with no need for extra brackets, because the horizontal line on the top of the square-root sign expands to cover the whole argument, such as in $\sqrt{2}$ vs. $\sqrt{22222}$. But still, you can write most of what you need in plain text; it just needs a tiny bit of extra work on your part.

I am sorry about the poor plain text representation. When I was writing this I was on time constraints due to class, but I don't understand why helpers are sensitive to written work. I get that some can be hard and such, but I guarantee there is research showing that writing is better to learn than typing. Yes, I could just do it twice by writing it and then typing, but that just increases the amount of time I have to take out to do so, which unfortuanatley I do not have. I understand that you guys too are under time constraints so to make this easier I will learn the commands and make it look better as well as clear. I simply did not have time this morning to do that though.

 

[Ray Vickson]

I am sorry about the poor plain text representation. When I was writing this I was on time constraints due to class, but I don't understand why helpers are sensitive to written work. I get that some can be hard and such, but I guarantee there is research showing that writing is better to learn than typing. Yes, I could just do it twice by writing it and then typing, but that just increases the amount of time I have to take out to do so, which unfortuanatley I do not have. I understand that you guys too are under time constraints so to make this easier I will learn the commands and make it look better as well as clear. I simply did not have time this morning to do that though.

OK. Back to your questions.
For series 1, you said you had no problem doing the ratio test. However, you said you could not get the ratio test to work for series 2. I don't understand that: series 1 and 2 are very similar, and what works for one ought to work for the other.

Another way of dealing with series 2 is to give an explicit, closed-form expression for the finite sum$\sum_{k=0}^N k (2/3)^k,$ then see if it has a limit when $N \to \infty$.

 

[mfb]

The third series I just compared 1/n² and got taht it converges

I see no interpretation of Σ√n/1+n² where such a comparison would work.

$\frac{\sqrt n}{1+n^2} > \frac 1 {n^2}$ and$\sqrt { \frac{n}{1+n^2}} > \frac 1 {n^2}$ and $\frac{\sqrt n}{1}+n^2 > \frac 1 {n^2}$ for almost all n.
Which interpretation is your problem?

Same question for the fourth series.If you would have used the time complaining about LaTeX to learn the basic commands, you would know them by now.

 

[Mark44]

If you would have used the time complaining about LaTeX to learn the basic commands, you would know them by now.

The same thought occurred to me.

 

[Ray Vickson]

I see no interpretation of Σ√n/1+n² where such a comparion would work.

$\frac{\sqrt n}{1+n^2} > \frac 1 {n^2}$ and$\sqrt { \frac{n}{1+n^2}} > \frac 1 {n^2}$ and $\frac{\sqrt n}{1}+n^2 > \frac 1 {n^2}$ for almost all n.
Which interpretation is your problem?

Same question for the fourth series.If you would have used the time complaining about LaTeX to learn the basic commands, you would know them by now.

He make it more-or-less clear that he objects to having to type the work, so I guess that not even using parentheses in plain text will work for him. I, for one, am unwilling to spend any more time trying to help him

 

[jlmccart03]

I see no interpretation of Σ√n/1+n² where such a comparison would work.

$\frac{\sqrt n}{1+n^2} > \frac 1 {n^2}$ and$\sqrt { \frac{n}{1+n^2}} > \frac 1 {n^2}$ and $\frac{\sqrt n}{1}+n^2 > \frac 1 {n^2}$ for almost all n.
Which interpretation is your problem?

Same question for the fourth series.If you would have used the time complaining about LaTeX to learn the basic commands, you would know them by now.

The same thought occurred to me.

He make it more-or-less clear that he objects to having to type the work, so I guess that not even using parentheses in plain text will work for him. I, for one, am unwilling to spend any more time trying to help him

I don't get all the hostitlity? Seriously, I hope to god you guys do not work with students in person because THIS is not how you teach. Period. You navigate to the students needs, not the other way around. I took the time this weekend to learn the commands, but I DID NOT HAVE TIME ON FRIDAY. I complained because I didn't understand why this site has issues with simple things such as photos and written work. IF you are a teacher then answer this: How do you deal with students in a classroom with poor handwriting? I get you guys have limited time, again, I understand this, but why become a mentor/advisor/etc if you guys are simply going to give me attitude? THIS right here is the reason American Education is in the crapper compared to many other countries. So thanks.

 

[mfb]

Most of my post was asking for clarification about your series and commenting on your approach to solve it.
I added a small side-remark about LaTeX because I didn't understand the point of the discussion here.

How do you deal with students in a classroom with poor handwriting?

If it is not legible, teachers everywhere around the world will deduct points or simply not count it. Even teachers who are paid to read it. Here no one is paid to help.

THIS right here is the reason American Education is in the crapper compared to many other countries.

Out of three who replied here, only one is from the US.

 

[Ray Vickson]

I don't get all the hostitlity? Seriously, I hope to god you guys do not work with students in person because THIS is not how you teach. Period. You navigate to the students needs, not the other way around. I took the time this weekend to learn the commands, but I DID NOT HAVE TIME ON FRIDAY. I complained because I didn't understand why this site has issues with simple things such as photos and written work. IF you are a teacher then answer this: How do you deal with students in a classroom with poor handwriting? I get you guys have limited time, again, I understand this, but why become a mentor/advisor/etc if you guys are simply going to give me attitude? THIS right here is the reason American Education is in the crapper compared to many other countries. So thanks.

In my second response (to your question of whether you can post an image of your work) I said "please don't".

Nothing forbids you from posting an image; it is just the case that maybe nobody will read it. As I said in post #9, "We really discourage that; most helpers will not look at pictures of work, but you might get lucky and encounter one helper who is willing to overlook the issue." So, that leaves it up to you: submit an image and hope somebody reads it, or (if it is particularly legible and well-written) a helper who normally would not look at it might in this case. As a rule, I never look at messy, sideways images, but sometimes DO respond to easily-read and easily-deciphered ones.

None of that detracts from the fact that the de-facto standard for submitting work here is to type it out, and there are good reasons for that---it is not some arbitrary, dictatorial limitation. You asked about that, and then seemed to be offended by the answer, using such language as "Sorry to inconvienance you, but I cannot and will not type all of my work when I do not learn that way."

Meanwhile, you were not asking any questions about the math itself.

 

[jlmccart03]

In my second response (to your question of whether you can post an image of your work) I said "please don't". Nothing forbids you from posting an image; it is just the case that maybe nobody will read it. As I said in post #9, "We really discourage that; most helpers will not look at pictures of work, but you might get lucky and encounter one helper who is willing to overlook the issue." So, that leaves it up to you: submit an image and hope somebody reads it, or (if it is particularly legible and well-written) a helper who normally would not look at it might in this case. As a rule, I never look at messy, sideways images, but sometimes DO respond to easily-read and easily-deciphered ones.

None of that detracts from the fact that the de-facto standard for submitting work here is to type it out, and there are good reasons for that---it is not some arbitrary, dictatorial limitation. You asked about that, and then seemed to be offended by the answer, using such language as "Sorry to inconvienance you, but I cannot and will not type all of my work when I do not learn that way."

Meanwhile, you were not asking any questions about the math itself.

Ok, I hope to start over and more importantly make sure I understand the math behind these series. Here is what I have $$\sum_{n=0}^\infty {\frac{n^3} {3^n}}$$ This simply can be determined by using the ratio test to get $$\lim_{n \rightarrow +\infty} {\frac {n^3} {3^n}} = {\frac {1} {3}}$$ So this is less than 1 and therefore converges absolutley since it was done using the ratio test.

For the second series $$\sum_{n=0}^\infty k{\frac{2} {3}^k}$$ I used the ratio test also and got {\frac {2} {3}} as the answer making it converge absolutely.

For the third series $$\sum_{n=0}^\infty {\frac{\sqrt n} {1+n^2}}$$ I did limit comparison to $$n^{\frac {3}{2}}$$ and got the answer to 1 so converges since my $$b_n = \frac {1} {n^{\frac 3 2}}$$ converges too.

Finally for the last series $$\sum_{n=0}^\infty (-1)^{n+1}{\frac{n} {n^2+9}}$$ it converges conditionally as I used alternating series test and found that the n+1 is less than the comparing piece $${\frac{n} {n^2+9}}$$ and the limit was 0 so it converges, but do limit comparison I find that the limit is not 0 and thus diverges so it is conditionally.

 

[Mark44]

Thank you for making the effort that is evident in this post! We appreciate it.

Ok, I hope to start over and more importantly make sure I understand the math behind these series. Here is what I have $$\sum_{n=0}^\infty {\frac{n^3} {3^n}}$$ This simply can be determined by using the ratio test to get $$\lim_{n \rightarrow +\infty} {\frac {n^3} {3^n}} = {\frac {1} {3}}$$ So this is less than 1 and therefore converges absolutley since it was done using the ratio test.

Looks good.

For the second series $$\sum_{n=0}^\infty k{\frac{2} {3}^k}$$ I used the ratio test also and got {\frac {2} {3}} as the answer making it converge absolutely.

Based on post #1, the series is $\sum_{n=0}^\infty k (\frac{2} {3})^k$, which converges just as you said.

For the third series $$\sum_{n=0}^\infty {\frac{\sqrt n} {1+n^2}}$$ I did limit comparison to $$n^{\frac {3}{2}}$$ and got the answer to 1 so converges since my $$b_n = \frac {1} {n^{\frac 3 2}}$$ converges too.

Yep. And you made the best choice of a series to compare to.

Finally for the last series $$\sum_{n=0}^\infty (-1)^{n+1}{\frac{n} {n^2+9}}$$ it converges conditionally as I used alternating series test and found that the n+1 is less than the comparing piece $${\frac{n} {n^2+9}}$$ and the limit was 0 so it converges, but do limit comparison I find that the limit is not 0 and thus diverges so it is conditionally.

Just right. The alternating series converges, but the series $\sum \frac n {n^2 + 1}$ diverges (by comparison to the harmonic series $\sum \frac 1 n$).
I'm not sure I understand what you're saying in "I used alternating series test and found that the n+1 is less than the comparing piece..." For the alt. series test, the terms have to be decreasing and $\lim a_n$ has to be 0 (excluding the sign changes by (-1)^{n + 1} in both). I think that in "n+1 is less than ..." you're saying that the a_n's are decreasing.

 

[jlmccart03]

Thank you for making the effort that is evident in this post! We appreciate it.
Looks good.
Based on post #1, the series is $\sum_{n=0}^\infty k (\frac{2} {3})^k$, which converges just as you said.

Yep. And you made the best choice of a series to compare to.
Just right. The alternating series converges, but the series $\sum \frac n {n^2 + 1}$ diverges (by comparison to the harmonic series $\sum \frac 1 n$).
I'm not sure I understand what you're saying in "I used alternating series test and found that the n+1 is less than the comparing piece..." For the alt. series test, the terms have to be decreasing and $\lim a_n$ has to be 0 (excluding the sign changes by (-1)^{n + 1} in both). I think that in "n+1 is less than ..." you're saying that the a_n's are decreasing.

Yes, for the last series I was taught last week that they must be of decreasing terms in the sense of 0 < a_n+1 < a_n. I wrote it in a different way that came off confusing so sorry about that. Thanks for staying with me on this, I realize now that it would have been just simpler to learn and post through typing. Now that I know them I will be able to make future post much more clear to the reader!

 

[Mark44]

Good. You've made a lot of progress on using TeX. Here are a couple of tips, mostly on the use of the braces.
Exponents
If an exponent is a single character or single digit, you don't need braces, e.g. ##x^3##, which renders as $x^3$. However, if the exponent is two or more characters, you have to use braces around them, ##x^{-3}##, rendering as $x^{-3}$. Without the braces, you get $x^-3$, where the sign is an exponent, but the 3 isn't.
Fractions
The same idea holds with fractions -- if both the numerator and denominator are single characters, you don't need the braces. E.g., ##\frac 2 3## ($\frac 2 3$), but 2x/(x + y) would need to be written as ##\frac{2x}{x + y}##.
Summations
\sum_{n = 1}^{\infty} \frac 2 n
You don't need braces around the expression in your summation, but you do need them for the starting value of your summation index, and possibly for the upper limit. For \infty, apparently you don't need braces around it, as I saw in your writing.

Inline vs Standalone
Everything I wrote above is done as inline TeX, using a pair of ## characters at each end. With inline TeX you can insert what you write seamlessly into the rest of your text. Standalone TeX uses a pair of $$ characters at each end, as you know. Stuff written as standalone TeX will appear centered on its own line.

 

[Ray Vickson]

Ok, I hope to start over and more importantly make sure I understand the math behind these series. Here is what I have $$\sum_{n=0}^\infty {\frac{n^3} {3^n}}$$ This simply can be determined by using the ratio test to get $$\lim_{n \rightarrow +\infty} {\frac {n^3} {3^n}} = {\frac {1} {3}}$$ So this is less than 1 and therefore converges absolutley since it was done using the ratio test.

For the second series $$\sum_{n=0}^\infty k{\frac{2} {3}^k}$$ I used the ratio test also and got {\frac {2} {3}} as the answer making it converge absolutely.

For the third series $$\sum_{n=0}^\infty {\frac{\sqrt n} {1+n^2}}$$ I did limit comparison to $$n^{\frac {3}{2}}$$ and got the answer to 1 so converges since my $$b_n = \frac {1} {n^{\frac 3 2}}$$ converges too.

Finally for the last series $$\sum_{n=0}^\infty (-1)^{n+1}{\frac{n} {n^2+9}}$$ it converges conditionally as I used alternating series test and found that the n+1 is less than the comparing piece $${\frac{n} {n^2+9}}$$ and the limit was 0 so it converges, but do limit comparison I find that the limit is not 0 and thus diverges so it is conditionally.

For series 1: OK. For series 2, OK if you mean $k (2/3)^k$ (rather than the $k 2^k / 3$ that you wrote). For series 3: OK. Finally, for series 4: OK as well. BTW: for an alternating series we just need $|a_{n+1}| < |a_n|$ eventually (that is, for all $n$ larger than some finite $N$), so even if you do not have that condition for the first hundred terms, as long as you have it from some point onward the alternating series theorem holds. In other words, do not be confused by a series like $1 - 2 + 2 - 3 + 3.2 - 1.5 + 1 - .75 + .5 - \cdots$, whose first few terms do not obey the "decreasing"condition.