Series: Determine if they are convergent or divergent

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1. Mar 10, 2017

jlmccart03

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Comparison Test
Ratio Test
Alternating Series Test
Divergence Test, etc

3. 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 an < bn (where an is the third series in the bullet point list and bn 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.

2. Mar 10, 2017

Staff: Mentor

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?

3. Mar 10, 2017

Ray Vickson

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.

4. Mar 10, 2017

jlmccart03

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

5. Mar 10, 2017

Staff: Mentor

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

6. Mar 10, 2017

jlmccart03

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.

7. Mar 10, 2017

Staff: Mentor

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.

8. Mar 10, 2017

jlmccart03

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.

9. Mar 10, 2017

Ray Vickson

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.

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.

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.

Last edited: Mar 11, 2017
10. Mar 10, 2017

Staff: Mentor

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

11. Mar 10, 2017

jlmccart03

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.

12. Mar 10, 2017

Ray Vickson

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 $N \to \infty$.

13. Mar 11, 2017

Staff: Mentor

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.

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.

14. Mar 11, 2017

Staff: Mentor

The same thought occurred to me.

15. Mar 11, 2017

Ray Vickson

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

16. Mar 12, 2017

jlmccart03

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.

17. Mar 12, 2017

Staff: Mentor

I added a small side-remark about LaTeX because I didn't understand the point of the discussion here.

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.

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

18. Mar 12, 2017

Ray Vickson

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."

Last edited: Mar 12, 2017
19. Mar 12, 2017

jlmccart03

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.

Last edited by a moderator: Mar 12, 2017
20. Mar 12, 2017

Staff: Mentor

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 an's are decreasing.