Test series for convergence

In summary: For (b), it's possible that the series converges by a different method.For (c), the sequence is too complicated to use the squeeze theorem.For (d), the sequence converges by the limit comparison test.For (e), the limit comparison test is inconclusive.For (f), the series converges by the ratio test.
  • #1
DryRun
Gold Member
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4

Homework Statement


[tex](a)\;\sum^{\infty}_{n=1}\frac{n-5}{n^2}\;(solved)[/tex]
[tex](b)\;\sum^{\infty}_{r=1}\frac{2r}{1+r^2}\;(solved)[/tex]
[tex](c)\;\sum^{\infty}_{n=1}\frac{\cos^4 nx}{n^2}\;(solved)[/tex]
[tex](d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}[/tex]
[tex](e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}\;(solved)[/tex]
[tex](f)\;\sum^{\infty}_{n=1}\frac{1.2.3...n}{4.7.10... (3n+1)}\;(solved)[/tex]

Homework Equations


nth-term test, ratio test, comparison test, limit comparison test, geometric series, etc.

The Attempt at a Solution


[tex](a)\;\sum^{\infty}_{n=1}\frac{n-5}{n^2}=\sum^{\infty}_{n=1}\frac{1}{n}-\sum^{\infty}_{n=1}\frac{5}{n^2}[/tex]This is contradictory, as according to the p-series test, the first series diverges but the second series converges. I also tried to factorize out ##n## or ##n^2## out of the numerator and denominator, but got limit = 0, which is inconclusive.

[tex](b)\;\lim_{r\to \infty}\frac{2}{1/r+r}=0[/tex]I factorized ##r## out of the numerator and denominator, then took the limit but got 0, which is inconclusive. I also tried comparison test, since ##v_n=2/n \ge u_n## and the series ##v_n## diverges, the test is inconclusive.

[tex](c)\;0\le \frac{\cos^4 nx}{n^2}\le 1/n^2[/tex]
I used the squeeze theorem on the sequence, but got 0, which is inconclusive.

[tex](d)\;\lim_{n\to \infty}\frac{1+(3/4)^r}{1+(5/4)^r}=1/∞=0[/tex]
Since the limit = 0, it is inconclusive.

[tex](e)\;\lim_{r\to \infty}\left( \frac{r+1}{r} \right)^r[/tex]
I used the ratio test. Then got stuck. Took logs and got the limit = ##e^∞##?

[tex](f)\;\lim_{n\to \infty}\frac{(4+4/n)}{(3+4/n)}=4/3[/tex]
I used the ratio test. L>1, so the series diverges. But the answer is convergent series.
 
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  • #2
(a), (b) Which tests apply? Which have you tried?

(c) How is the squeeze theorem inconclusive?

(d)-(e) These are sequences. The tests for convergence/divergence (which are for series) do not apply.

Edit: Wow sorry. I really need to look at the whole problem.
 
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  • #3
(c) Looks like you have a pretty good setup for using one of the tests (not the test for divergence).

(d) I'd split the ratio into two pieces and examine them individually.

(e) This is a very common limit. I guarantee you've seen it at least two or three times in your calculus career.

(f) I'd check for an algebra mistake. You've got the right idea, though.
 
  • #4
For (c): Use the comparison test. Since the original series is less than ##1/n^2## and the latter converges, therefore the original series converges.
[tex](d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}= \sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}+ \sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}[/tex]The limits of both sequences give: 1/(∞+∞) + 1/∞ = 0 and the result is inconclusive.

For (f): Using the ratio test, here is what i end up with: [tex]\frac{u_{n+1}}{u_n}=\frac{4n+4}{3n+4}[/tex]I get: L = 4/3
 
  • #5
Comparison test is very good for most of these series :smile:

For a), after splitting the series up like that and showing that part of the series diverges, the original series diverges too.

For b), try starting with 1+r2 < 2r2

For d), start with 4r + 5r > 5r

For e), what's the limit of the sequence?

f) is a little more complicated than how you worked it out, since all those numbers are multiplied together and you're not just considering n and (3n+1). Still thinking about that one...
 
  • #6
There are finitely many (seven by my count) tests:

The nth-term test rarely gives you anything conclusive, but it's usually easy to compute. In the problems you run across in calculus classes, it almost never "works". You can't just give up when it's inconclusive, because it almost always will be.

The integral test and the alternating series test usually don't apply due to the conditions you need in order to use them. When they do apply, they give you definitive answers and you're done.

Ratio/root are algorithmic; once you decide to try, you just do it. Either you get something conclusive or you don't.

So really you only have two tests where any decisions need to be made; comparison and limit comparison. Protip: unless the direct comparison is obvious (like it was in c above) use the limit comparison. It's way easier.

So at the risk of sounding mean, do the flipping tests. There's no pressure time wise right now. Just do them. Figure out which ones work in certain cases and which ones don't. That kind of intuition only comes with practice. So, again, just do it.

P.S. When all of your efforts seem to be in vain, make sure you haven't got a telescoping series sitting in front of you.
 
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  • #7
It often helps to look at how the terms behave far out in the sequence. For example, for (a), when n gets large, the terms look like
$$\frac{n-5}{n^2} \approx \frac{n}{n^2} = \frac{1}{n}$$ This suggests the series will diverge. To prove it, try using the limit comparison test with 1/n.
 
  • #8
Depending on whether you think the sum in f) converges or diverges, you can set up an inequality with 4·7·10···(3n+1) and 3·6·9···(3n) or 6·9·12···(3n+3).
 
  • #9
gopher_p said:
There are finitely many (seven by my count) tests:

The nth-term test rarely gives you anything conclusive, but it's usually easy to compute. In the problems you run across in calculus classes, it almost never "works". You can't just give up when it's inconclusive, because it almost always will be.

The integral test and the alternating series test usually don't apply due to the conditions you need in order to use them. When they do apply, they give you definitive answers and you're done.

Ratio/root are algorithmic; once you decide to try, you just do it. Either you get something conclusive or you don't.

So really you only have two tests where any decisions need to be made; comparison and limit comparison. Protip: unless the direct comparison is obvious (like it was in c above) use the limit comparison. It's way easier.

So at the risk of sounding mean, do the flipping tests. There's no pressure time wise right now. Just do them. Figure out which ones work in certain cases and which ones don't. That kind of intuition only comes with practice. So, again, just do it.

P.S. When all of your efforts seem to be in vain, make sure you haven't got a telescoping series sitting in front of you.
I'll be sure to keep that in mind.
Bohrok said:
Depending on whether you think the sum in f) converges or diverges, you can set up an inequality with 4·7·10···(3n+1) and 3·6·9···(3n) or 6·9·12···(3n+3).
I used the ratio test and i am quite sure that the limit L = 4/3
vela said:
It often helps to look at how the terms behave far out in the sequence. For example, for (a), when n gets large, the terms look like
$$\frac{n-5}{n^2} \approx \frac{n}{n^2} = \frac{1}{n}$$ This suggests the series will diverge. To prove it, try using the limit comparison test with 1/n.
Thanks for the help, vela. I tried with comparison test but the result was inconclusive, then i tried with the limit comparison test and got the answer = 1. Since ##0≤L≤∞## therefore either both series converge or both diverge. Since 1/n diverges, therefore the original series diverges. I used the same method to solve (b).
 
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  • #10
sharks said:
For (f): Using the ratio test, here is what i end up with: [tex]\frac{u_{n+1}}{u_n}=\frac{4n+4}{3n+4}[/tex]I get: L = 4/3
Show us how you got that ratio. I got something different and found the series converges.
 
  • #11
vela said:
Show us how you got that ratio. I got something different and found the series converges.
For part (f): using the ratio test
[tex]u_n=\frac{1.2.3...n}{4.7.10... (3n+1)}
\\u_{n+1}=\frac{2.3.4...(n+1)}{7.10.13... (3n+4)}
\\\frac{u_{n+1}}{u_n}=\frac{2.3.4...(n+1)}{7.10.13... (3n+4)}\times \frac{4.7.10... (3n+1)}{1.2.3...n}=\frac{4(n+1)}{(3n+4)}=\frac{4n+4}{3n+4}
\\\lim_{n \to \infty}\frac{4n+4}{3n+4}=4/3[/tex]
 
  • #12
Your expression for un+1 isn't correct. You shouldn't be dropping the first factor in the numerator and denominator.
 
  • #13
vela said:
Your expression for un+1 isn't correct. You shouldn't be dropping the first factor in the numerator and denominator.
[tex]u_n=\frac{1.2.3...n}{4.7.10... (3n+1)}
\\u_{n+1}=\frac{1.2.3.4...(n+1)}{4.7.10.13... (3n+4)}
\\\frac{u_{n+1}}{u_n}=\frac{1.2.3.4...(n+1)}{4.7.10.13... (3n+4)}\times \frac{4.7.10... (3n+1)}{1.2.3...n}=\frac{(n+1)}{(3n+4)}
\\\lim_{n \to \infty}\frac{n+1}{3n+4}=\frac{1}{3}[/tex]Since L < 1, the original series converges.
 
  • #14
[tex](e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}[/tex]Using the ratio test:
[tex]u_n=\frac{r^r}{r!}
\\u_{n+1}=\frac{(r+1)^{r+1}}{(r+1)!}=\frac{(r+1)^r.(r+1)}{r!(r+1)}
\\\frac{u_{n+1}}{u_n}=\frac{(r+1)^r.(r+1)}{r!(r+1)}\times \frac{r!}{r^r}=\frac{(r+1)^r}{r^r}=\left( \frac{r+1}{r} \right)^r
\\\lim_{n\to \infty} \frac{u_{n+1}}{u_n}=\lim_{n\to \infty}\left( \frac{r+1}{r} \right)^r
\\y=\left( \frac{r+1}{r} \right)^r
\\\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}=-r^3/(r+1)
\\y=e^{-r^3/(r+1)}
\\\lim_{n\to \infty}e^{-r^3/(r+1)}=\lim_{n\to \infty}e^{-\infty}=-∞
[/tex]Since L=-∞, meaning L<1, therefore the original series converges. Is that correct?
 
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  • #15
sharks said:
[tex](e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}[/tex]Using the ratio test:
[tex]u_n=\frac{r^r}{r!}
\\u_{n+1}=\frac{(r+1)^{r+1}}{(r+1)!}=\frac{(r+1)^r.(r+1)}{r!(r+1)}
\\\frac{u_{n+1}}{u_n}=\frac{(r+1)^r.(r+1)}{r!(r+1)}\times \frac{r!}{r^r}=\frac{(r+1)^r}{r^r}=\left( \frac{r+1}{r} \right)^r
\\\lim_{n\to \infty} \frac{u_{n+1}}{u_n}=\lim_{n\to \infty}\left( \frac{r+1}{r} \right)^r
\\y=\left( \frac{r+1}{r} \right)^r
\\\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}=-r^3/(r+1)
\\y=e^{-r^3/(r+1)}
\\\lim_{n\to \infty}e^{-r^3/(r+1)}=\lim_{n\to \infty}e^{-\infty}=-∞
[/tex]Since L=-∞, meaning L<1, therefore the original series converges. Is that correct?

I don't think so. I had trouble spotting your error, but lim[itex]_{r\rightarrow\infty}[/itex]=[itex]\infty[/itex], so the series should diverge.
 
  • #16
sharks said:
[tex]\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}[/tex]
You're fine up to here. Simplify the argument of the log to get
$$\lim_{r \to \infty} \frac{\log \big(1+\frac{1}{r}\big)}{1/r}$$ and try evaluating it again using the Hospital rule.

Since L=-∞, meaning L<1, therefore the original series converges. Is that correct?
That limit can't possibly be right. The function you're taking the limit of is positive. How can you end up with a negative limit?
 
  • #17
vela said:
You're fine up to here. Simplify the argument of the log to get
$$\lim_{r \to \infty} \frac{\log \big(1+\frac{1}{r}\big)}{1/r}$$ and try evaluating it again using the Hospital rule.
[tex]\\\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}=\frac{\ln \left( 1+\frac{1}{r} \right)}{1/r}[/tex]Now, using L'Hopital's rule:[tex]\lim_{r\to \infty}\ln y=\frac{1/ \left( 1+\frac{1}{r} \right)}{-1/r^2}=\frac{-r^2}{ \left( 1+\frac{1}{r} \right)}
\\\ln \lim_{r\to \infty}y=\frac{-r^2}{ \left( 1+\frac{1}{r} \right)}
\\\lim_{r\to \infty}y=e^{\frac{-r^2}{ \left( 1+\frac{1}{r} \right)}}=e^{-∞}[/tex]Is that correct? I had to analyze it on graph to confirm but i think that as ##r→∞, e^{-∞}=0##? Since L < 1, the original series converges? But the answer is divergent series.
 
  • #18
You need to use the chain rule when you differentiate the logarithm.

Again, the negative result should have set off alarms that you made a mistake. ln(1+1/r) > 0 and 1/r>0. How can the limit diverge to -∞?
 
  • #19
Using L'Hopital's rule:[tex]\lim_{r\to \infty}\ln y=\frac{1/ \left( 1+\frac{1}{r} \right).(-1/r^2)}{-1/r^2}=\frac{1}{ \left( 1+\frac{1}{r} \right)}
\\\ln \lim_{r\to \infty}y=\frac{1}{ \left( 1+\frac{1}{r} \right)}=\frac{r}{r+1}=1+\frac{1}{r}
\\\lim_{r\to \infty}y=e^{1+\frac{1}{r}}=e^1[/tex]Since L >1, the original series diverges.
 
  • #20
sharks said:
[tex](d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}= \sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}+ \sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}[/tex]The limits of both sequences give: 1/(∞+∞) + 1/∞ = 0 and the result is inconclusive.
I have no idea what to do next.
 
  • #21
Part (d): After struggling with this for a while, I've used the comparison test on each series:

Comparing with ##v_n=1/r##, since ##v_n\le u_n## and ##v_n## diverges, therefore both series diverge, and the sum of two divergent series is a divergent series.

This is the best that i could come up with. However, the answer says the series is convergent. I'm stuck.
 
  • #22
sharks said:
Part (d): After struggling with this for a while, I've used the comparison test on each series:

Comparing with ##v_n=1/r##, since ##v_n\le u_n## and ##v_n## diverges, therefore both series diverge, and the sum of two divergent series is a divergent series.

This is the best that i could come up with. However, the answer says the series is convergent. I'm stuck.

Try using the ratio test for part (d).
 
  • #23
Infinitum said:
Try using the ratio test for part (d).

Hi Infinitum

[tex](d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}= \sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}+ \sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}[/tex]
Using ratio test on:[tex]\sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}[/tex]
[tex]\frac{u_{n+1}}{u_n}=\frac{3.4^r+3.5^r}{4.4^r+5.5^r}=3/5[/tex]Hence the series converges.
For the other series:[tex]\sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}[/tex]
[tex]\frac{u_{n+1}}{u_n}= \frac{4}{5}[/tex]Hence the series converges.

The sum of two convergent series is a convergent series.
 
  • #24
sharks said:
Hi Infinitum

[tex](d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}= \sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}+ \sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}[/tex]
Using ratio test on:[tex]\sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}[/tex]
[tex]\frac{u_{n+1}}{u_n}=\frac{3.4^r+3.5^r}{4.4^r+5.5^r}=3/5[/tex]Hence the series converges.
For the other series:[tex]\sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}[/tex]
[tex]\frac{u_{n+1}}{u_n}= \frac{4}{5}[/tex]Hence the series converges.

The sum of two convergent series is a convergent series.

Yep! :approve:

You could have done this without breaking up the original limit, too :wink:
 
  • #25
Infinitum said:
You could have done this without breaking up the original limit, too :wink:

I used the ratio test directly on the original series but it gets complicated and there is factor multiplication involved (as none of the terms cancel out) in both the numerator and denominator which expands so I'm not sure if it resolves to some value.
 
  • #26
sharks said:
I used the ratio test directly on the original series but it gets complicated and there is factor multiplication involved (as none of the terms cancel out) in both the numerator and denominator which expands so I'm not sure if it even resolves to some value.

Actually, you're right. I ended splitting it up at a later stage, probably where you got stuck.
 
  • #27
Hope not to throw off the subject too much; just curious as to how testing convergence

of series became important to Calculus/Analysis. Is there something other than a

connection with Riemann sums?
 
  • #28
You need to clean up your notation.
sharks said:
[tex]\lim_{r\to \infty}\ln y=\frac{1/ \left( 1+\frac{1}{r} \right).(-1/r^2)}{-1/r^2}=\frac{1}{ \left( 1+\frac{1}{r} \right)}[/tex]
Should be
$$\lim_{r\to\infty}\ln y
= \lim_{r\to\infty} \frac{1/ \left( 1+\frac{1}{r} \right)\cdot(-1/r^2)}{-1/r^2}
= \lim_{r\to\infty} \frac{1}{1+\frac{1}{r}} = 1
$$ You need to include ##\displaystyle \lim_{r\to\infty}## until you actually evaluate the limit.

Same problem with
$$\ln \lim_{r\to \infty}y=\frac{1}{ \left( 1+\frac{1}{r} \right)}=\frac{r}{r+1}=1+\frac{1}{r} \\
\lim_{r\to \infty}y=e^{1+\frac{1}{r}}=e^1$$
To say ##\ln \displaystyle\lim_{r\to \infty}y## equals ##\frac{1}{ \left( 1+\frac{1}{r} \right)}## is wrong. The variable r won't appear in the result of the limit, yet what you've written says that the limit is equal to an expression that depends on r.

The sequence also strikes me as being a bit disjointed from what you wrote above. You found the limit of ln y, and then now you're talking about the log of the limit of y. Just for clarity, you might want to add the step
$$\ln \lim_{r\to\infty} y = \lim_{r\to\infty} \ln y = \cdots$$ so the grader can tell you know what you're doing. Alternatively, you could have also said
$$\lim_{r\to\infty} y = \lim_{r\to\infty} e^{\ln y} = \exp\left(\lim_{r\to\infty} \ln y\right) = e^1 = e$$
 
  • #29
vela said:
You need to clean up your notation.

Should be
$$\lim_{r\to\infty}\ln y
= \lim_{r\to\infty} \frac{1/ \left( 1+\frac{1}{r} \right)\cdot(-1/r^2)}{-1/r^2}
= \lim_{r\to\infty} \frac{1}{1+\frac{1}{r}} = 1
$$ You need to include ##\displaystyle \lim_{r\to\infty}## until you actually evaluate the limit.

Same problem with

To say ##\ln \displaystyle\lim_{r\to \infty}y## equals ##\frac{1}{ \left( 1+\frac{1}{r} \right)}## is wrong. The variable r won't appear in the result of the limit, yet what you've written says that the limit is equal to an expression that depends on r.

The sequence also strikes me as being a bit disjointed from what you wrote above. You found the limit of ln y, and then now you're talking about the log of the limit of y. Just for clarity, you might want to add the step
$$\ln \lim_{r\to\infty} y = \lim_{r\to\infty} \ln y = \cdots$$ so the grader can tell you know what you're doing. Alternatively, you could have also said
$$\lim_{r\to\infty} y = \lim_{r\to\infty} e^{\ln y} = \exp\left(\lim_{r\to\infty} \ln y\right) = e^1 = e$$

Thanks for the advice, vela. Actually, i have my finals coming up very soon, and i agree, my presentation of the limits isn't completely correct.
 

1. What is a test series for convergence?

A test series for convergence is a mathematical tool used to determine whether an infinite series converges or diverges. It involves evaluating the limit of the series as the number of terms approaches infinity.

2. How do you determine if a series converges using a test series?

There are several different test series that can be used to determine convergence, such as the ratio test, the root test, and the integral test. Each test has specific criteria that must be met for a series to converge.

3. What is the difference between a convergent and divergent series?

A convergent series is one in which the sum of all the terms approaches a finite value as the number of terms increases. A divergent series, on the other hand, is one in which the sum of the terms either increases without bound or oscillates between positive and negative values.

4. Can a series converge to more than one value?

No, a series can only converge to a single value. If the sum of the terms approaches more than one value, then the series is said to be divergent.

5. Why is it important to determine convergence of a series?

Determining convergence is important because it allows us to understand the behavior of an infinite series and make predictions about its sum. It also has applications in various fields such as physics, engineering, and finance.

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