(adsbygoogle = window.adsbygoogle || []).push({}); 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.

[tex](a)\;\sum^{\infty}_{n=0}\frac{1}{3^{n-1}}[/tex]

[tex](b)\;\sum^{\infty}_{n=0}\frac{4n^2+n}{(n^7-n^3)^{1/3}}[/tex]

[tex](c)\;\sum^{\infty}_{n=0}\frac{\sin^2n}{2^n}[/tex]

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:

[tex]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[/tex]

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.

[tex]0 \le \sin^2n\le 1

\\\frac{0}{2^n} \le \frac{\sin^2n}{2^n}\le \frac{1}{2^n}[/tex]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.

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# Homework Help: Find limits of series

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