# Find limits of series

1. May 19, 2012

### sharks

1. The problem statement, all variables and given/known data
Determine if the following series converges or diverges and find the limits for those series that converge.
$$(a)\;\sum^{\infty}_{n=0}\frac{1}{3^{n-1}}$$
$$(b)\;\sum^{\infty}_{n=0}\frac{4n^2+n}{(n^7-n^3)^{1/3}}$$
$$(c)\;\sum^{\infty}_{n=0}\frac{\sin^2n}{2^n}$$
2. Relevant equations
The nth-term test, ratio test, comparison test, geometric series, etc.

3. The attempt at a solution

For part (a): I used the ratio test and got the answer: 1/3 as the limit of the convergent series. However, the answer is supposed to be: 3/2?

For part (b): I used the comparison test. For n large, i used the following approximation:
$$v_n=\sum^{\infty}_{n=0}\frac{4n^2}{(n^7)^{1/3}} \\L=\lim_{n\to \infty}\frac{u_n}{v_n}=0\times {\infty}=0$$
Since $L = 0$ and $v_n$ converges according to the p-series test, therefore the original series converges and the limit of the series is zero.
Apparently, it's wrong, as the series is supposed to be divergent.

For part (c): I'm thinking about the squeeze theorem to find if the limit is not zero.
$$0 \le \sin^2n\le 1 \\\frac{0}{2^n} \le \frac{\sin^2n}{2^n}\le \frac{1}{2^n}$$I'm stuck.

Edit: OK, i think i figured out part (b): The approximation was wrong. I then used 4/n and got a divergent series.

Last edited: May 19, 2012
2. May 19, 2012

### Dustinsfl

For (a)

$$\sum_{n=1}^{\infty}\frac{1}{3^{n-1}} = 3\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n$$

The answer should be 9/2 not 3/2

Last edited: May 19, 2012
3. May 19, 2012

### sharks

So, it's a geometric series, with a = 3 and r = 1/3. The limit of the series becomes: 9/2?

But using the ratio test, i get the limit as 1/3? Or does the ratio test only say if the series is convergent or divergent? But, additionally, does it also give the limit of the series, as the value of L? If not, then in the limit comparison test, the value of L is also not the limit of the series??

4. May 19, 2012

### Dustinsfl

The rule is if L = 1 inconclusive, L < 1 converges, and L > 1 diverges.

5. May 19, 2012

### sharks

I know the value of L is used to determine if the series converges or not, but is that value of L also the limit of the original series?

6. May 19, 2012

### Dustinsfl

L only tells you those 3 things. It doesn't tell you what the value is.

7. May 19, 2012

### sharks

OK thanks, i was trying to get confirmation on this. I have a bunch of problems which might be wrong because of this wrong assumption.

Edit: Converting $\sin^2n=1/2 - \cos 2n /2$
Then, $$\sum^{\infty}_{n=0}\frac{\sin^2n}{2^n}= \sum^{\infty}_{n=0} \frac{1}{2^{n-1}} - \sum^{\infty}_{n=0}\frac{\cos 2n}{2^{n-1}}$$
But i'm not sure if this actually simplifies the problem.

Edit #2: OK, i'm trying to find another simpler way. How about factorizing $2^n$ out of the numerator and denominator? The limit gives 0 which isn't helpful.

Last edited: May 19, 2012
8. May 19, 2012

### Bohrok

If you had to find the sum Ʃ 1/2n, that's a geometric series (similar to the one in a) that converges, so the limit of 1/2n as n→∞ is 0. Now sin2n/2n is squeezed to 0

9. May 19, 2012

### sharks

Zero. But the middle term of the inequality can't vary from 0 to 0. It doesn't make sense, unless the middle term itself is 0?

10. May 19, 2012

### sharks

I'm not sure, as the limit of sin2n/2n = 0, doesn't say anything about the series converging or diverging, nor anything about the limit of the series.

Here is another try on this problem: as n approaches infinity, $0 \le \sin^2n\le 1$ becomes $sin^2n =1/2$, which is the average value.

Then, using the comparison test: $$u_n=\frac{\sin^2n}{2^n} \\v_n=\frac{1/2}{2^n}= \frac{1}{2^{n-1}} \\u_n \le v_n$$$v_n$ is a geometric series that converges to 1, therefore $u_n$ converges.

11. May 20, 2012

### Bohrok

It doesn't vary from 0 to 0. sin2n/2n is squeezed above and below to 0 only as n→∞.

You mean limit of the sequence, right? Someone may jump on you really hard if you don't specify whether you mean sequence or series.
Ʃ sin2n/2n converges since the sequence is less than 1/2n, whose sum converges.

That inequality isn't correct, plus $\frac{1}{n^{n-1}} = \frac{2}{2^n}, \text{not } \frac{1/2}{2^n}$; however if you had vn = 1/2n instead of 1/2n-1, that would converge. And since $$\sum\frac{1}{2^{n-1}} = 2\sum\frac{1}{2^n}$$ you could use that for your proof.