# Test series for convergence

1. May 21, 2012

### sharks

1. The problem statement, all variables and given/known data
$$(a)\;\sum^{\infty}_{n=1}\frac{n-5}{n^2}\;(solved)$$
$$(b)\;\sum^{\infty}_{r=1}\frac{2r}{1+r^2}\;(solved)$$
$$(c)\;\sum^{\infty}_{n=1}\frac{\cos^4 nx}{n^2}\;(solved)$$
$$(d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}$$
$$(e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}\;(solved)$$
$$(f)\;\sum^{\infty}_{n=1}\frac{1.2.3...n}{4.7.10... (3n+1)}\;(solved)$$

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

3. 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.

Last edited: May 22, 2012
2. May 21, 2012

### gopher_p

(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.

Last edited: May 21, 2012
3. May 21, 2012

### gopher_p

(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. May 21, 2012

### sharks

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

5. May 21, 2012

### Bohrok

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...

6. May 21, 2012

### gopher_p

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.

Last edited: May 21, 2012
7. May 21, 2012

### vela

Staff Emeritus
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. May 21, 2012

### Bohrok

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. May 21, 2012

### sharks

I'll be sure to keep that in mind.
I used the ratio test and i am quite sure that the limit L = 4/3
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).

Last edited: May 21, 2012
10. May 21, 2012

### vela

Staff Emeritus
Show us how you got that ratio. I got something different and found the series converges.

11. May 21, 2012

### sharks

For part (f): using the ratio test
$$u_n=\frac{1.2.3...n}{4.7.10... (3n+1)} \\u_{n+1}=\frac{2.3.4...(n+1)}{7.10.13... (3n+4)} \\\frac{u_{n+1}}{u_n}=\frac{2.3.4...(n+1)}{7.10.13... (3n+4)}\times \frac{4.7.10... (3n+1)}{1.2.3...n}=\frac{4(n+1)}{(3n+4)}=\frac{4n+4}{3n+4} \\\lim_{n \to \infty}\frac{4n+4}{3n+4}=4/3$$

12. May 21, 2012

### vela

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

13. May 21, 2012

### sharks

$$u_n=\frac{1.2.3...n}{4.7.10... (3n+1)} \\u_{n+1}=\frac{1.2.3.4...(n+1)}{4.7.10.13... (3n+4)} \\\frac{u_{n+1}}{u_n}=\frac{1.2.3.4...(n+1)}{4.7.10.13... (3n+4)}\times \frac{4.7.10... (3n+1)}{1.2.3...n}=\frac{(n+1)}{(3n+4)} \\\lim_{n \to \infty}\frac{n+1}{3n+4}=\frac{1}{3}$$Since L < 1, the original series converges.

14. May 22, 2012

### sharks

$$(e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}$$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?

Last edited: May 22, 2012
15. May 22, 2012

### the_kid

I don't think so. I had trouble spotting your error, but lim$_{r\rightarrow\infty}$=$\infty$, so the series should diverge.

16. May 22, 2012

### vela

Staff Emeritus
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.

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. May 22, 2012

### sharks

$$\\\ln y=\ln \left( \frac{r+1}{r} \right)^r=r\ln \left( \frac{r+1}{r} \right)=\frac{\ln \left( \frac{r+1}{r} \right)}{1/r}=\frac{\ln \left( 1+\frac{1}{r} \right)}{1/r}$$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.

18. May 22, 2012

### vela

Staff Emeritus
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. May 22, 2012

### sharks

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.

20. May 22, 2012

### sharks

I have no idea what to do next.