# Test series for convergence

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## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

$$(a)\;\sum^{\infty}_{n=1}\frac{n-5}{n^2}=\sum^{\infty}_{n=1}\frac{1}{n}-\sum^{\infty}_{n=1}\frac{5}{n^2}$$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.

$$(b)\;\lim_{r\to \infty}\frac{2}{1/r+r}=0$$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.

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

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

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

$$(f)\;\lim_{n\to \infty}\frac{(4+4/n)}{(3+4/n)}=4/3$$
I used the ratio test. L>1, so the series diverges. But the answer is convergent series.

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(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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(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.

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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.
$$(d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}= \sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}+ \sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}$$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: $$\frac{u_{n+1}}{u_n}=\frac{4n+4}{3n+4}$$I get: L = 4/3

Comparison test is very good for most of these series

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 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...

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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vela
Staff Emeritus
Homework Helper
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.

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).

Gold Member
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.
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
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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vela
Staff Emeritus
Homework Helper
For (f): Using the ratio test, here is what i end up with: $$\frac{u_{n+1}}{u_n}=\frac{4n+4}{3n+4}$$I get: L = 4/3
Show us how you got that ratio. I got something different and found the series converges.

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Show us how you got that ratio. I got something different and found the series converges.
For part (f): using the ratio test
$$u_n=\frac{1.2.3...n}{4.7.10... (3n+1)} \\u_{n+1}=\frac{2.3.4...(n+1)}{7.10.13... (3n+4)} \\\frac{u_{n+1}}{u_n}=\frac{2.3.4...(n+1)}{7.10.13... (3n+4)}\times \frac{4.7.10... (3n+1)}{1.2.3...n}=\frac{4(n+1)}{(3n+4)}=\frac{4n+4}{3n+4} \\\lim_{n \to \infty}\frac{4n+4}{3n+4}=4/3$$

vela
Staff Emeritus
Homework Helper
Your expression for un+1 isn't correct. You shouldn't be dropping the first factor in the numerator and denominator.

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Your expression for un+1 isn't correct. You shouldn't be dropping the first factor in the numerator and denominator.
$$u_n=\frac{1.2.3...n}{4.7.10... (3n+1)} \\u_{n+1}=\frac{1.2.3.4...(n+1)}{4.7.10.13... (3n+4)} \\\frac{u_{n+1}}{u_n}=\frac{1.2.3.4...(n+1)}{4.7.10.13... (3n+4)}\times \frac{4.7.10... (3n+1)}{1.2.3...n}=\frac{(n+1)}{(3n+4)} \\\lim_{n \to \infty}\frac{n+1}{3n+4}=\frac{1}{3}$$Since L < 1, the original series converges.

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$$(e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}$$Using the ratio test:
$$u_n=\frac{r^r}{r!} \\u_{n+1}=\frac{(r+1)^{r+1}}{(r+1)!}=\frac{(r+1)^r.(r+1)}{r!(r+1)} \\\frac{u_{n+1}}{u_n}=\frac{(r+1)^r.(r+1)}{r!(r+1)}\times \frac{r!}{r^r}=\frac{(r+1)^r}{r^r}=\left( \frac{r+1}{r} \right)^r \\\lim_{n\to \infty} \frac{u_{n+1}}{u_n}=\lim_{n\to \infty}\left( \frac{r+1}{r} \right)^r \\y=\left( \frac{r+1}{r} \right)^r \\\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}=-r^3/(r+1) \\y=e^{-r^3/(r+1)} \\\lim_{n\to \infty}e^{-r^3/(r+1)}=\lim_{n\to \infty}e^{-\infty}=-∞$$Since L=-∞, meaning L<1, therefore the original series converges. Is that correct?

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$$(e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}$$Using the ratio test:
$$u_n=\frac{r^r}{r!} \\u_{n+1}=\frac{(r+1)^{r+1}}{(r+1)!}=\frac{(r+1)^r.(r+1)}{r!(r+1)} \\\frac{u_{n+1}}{u_n}=\frac{(r+1)^r.(r+1)}{r!(r+1)}\times \frac{r!}{r^r}=\frac{(r+1)^r}{r^r}=\left( \frac{r+1}{r} \right)^r \\\lim_{n\to \infty} \frac{u_{n+1}}{u_n}=\lim_{n\to \infty}\left( \frac{r+1}{r} \right)^r \\y=\left( \frac{r+1}{r} \right)^r \\\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}=-r^3/(r+1) \\y=e^{-r^3/(r+1)} \\\lim_{n\to \infty}e^{-r^3/(r+1)}=\lim_{n\to \infty}e^{-\infty}=-∞$$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$_{r\rightarrow\infty}$=$\infty$, so the series should diverge.

vela
Staff Emeritus
Homework Helper
$$\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}$$
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?

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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.
$$\\\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}=\frac{\ln \left( 1+\frac{1}{r} \right)}{1/r}$$Now, using L'Hopital's rule:$$\lim_{r\to \infty}\ln y=\frac{1/ \left( 1+\frac{1}{r} \right)}{-1/r^2}=\frac{-r^2}{ \left( 1+\frac{1}{r} \right)} \\\ln \lim_{r\to \infty}y=\frac{-r^2}{ \left( 1+\frac{1}{r} \right)} \\\lim_{r\to \infty}y=e^{\frac{-r^2}{ \left( 1+\frac{1}{r} \right)}}=e^{-∞}$$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.

vela
Staff Emeritus
Homework Helper
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 -∞?

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Using L'Hopital's rule:$$\lim_{r\to \infty}\ln y=\frac{1/ \left( 1+\frac{1}{r} \right).(-1/r^2)}{-1/r^2}=\frac{1}{ \left( 1+\frac{1}{r} \right)} \\\ln \lim_{r\to \infty}y=\frac{1}{ \left( 1+\frac{1}{r} \right)}=\frac{r}{r+1}=1+\frac{1}{r} \\\lim_{r\to \infty}y=e^{1+\frac{1}{r}}=e^1$$Since L >1, the original series diverges.

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$$(d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}= \sum^{\infty}_{n=1}\frac{3^r}{4^r+5^r}+ \sum^{ \infty}_{n=1}\frac{4^r}{4^r+5^r}$$The limits of both sequences give: 1/(∞+∞) + 1/∞ = 0 and the result is inconclusive.
I have no idea what to do next.

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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.

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).

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Try using the ratio test for part (d).

Hi Infinitum

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

The sum of two convergent series is a convergent series.

Hi Infinitum

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

The sum of two convergent series is a convergent series.

Yep!

You could have done this without breaking up the original limit, too

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You could have done this without breaking up the original limit, too

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.