- #1
DryRun
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Homework Statement
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]
Homework Equations
The nth-term test, ratio test, comparison test, geometric series, etc.
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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