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I'm trying to show convergence/divergence of some series. The first is $\sum_{n=1}^\infty \frac{n}{n^2+2}$. First, I tried the comparison test predicting it was divergent as I thought it would behave like $\sum_{n=1}^\infty \frac{1}{n}$ so I tried to look for a 'smaller' series which was also divergent, but removing the 2 makes it larger so I could not find a way. I also tried the divergence test but the series terms go to 0 and the ratio test but that comes out as 1. I have not tried the integral test.

Have I done something wrong here?

Secondly, I can't do $\sum_{n=1}^\infty \frac{n^{100}}{10^n}$. I would predict this converges, I can't do the limit of the terms for the divergence test, tried the ratio test but can't do the resulting limit. And I can't see how to use the comparison test.

I appreciate any help thank you.

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    $\begingroup$ For the first one, replace the $2$ with a $2n^2$. $\endgroup$ – Clayton Jul 22 at 21:31
  • $\begingroup$ thanks, then what test would you use to show the resulting one is divergent? $\endgroup$ – Carlos Bacca Jul 22 at 21:33
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For the first one use limit comparison theorem.

For the second one redo the ratio test more carefully and it should work.

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(1) Consider the harmonic series (as you have), but in conjunction with the Limit Comparison test:

$$ 0 < \lim_{n \to \infty} \frac{\frac{n}{n^{2} + 2}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^{2}}{n^{2} + 2} = 1 < \infty.$$

Hence, since the harmonic series diverges, so does $\sum_{n=1}^{\infty} \frac{n}{n^{2}+2}$.

(2) Using the Ratio Test,

$$ \lim_{n \to \infty} \frac{(n+1)^{100}}{10^{n+1}} \cdot \frac{10^{n}}{n^{100}} = \lim_{n \to \infty} \frac{\left( \frac{n+1}{n} \right)^{100}}{10} = \frac{1}{10} < 1. $$

Therefore, $\sum_{n=1}^{\infty} \frac{n^{100}}{10^{n}}$ converges.

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For the second, one can use Cauchy criteria and compute $$\lim_{n\to+\infty} (\frac{n^{100}}{10^n})^{\frac 1n}$$

$$=\lim_{n\to\infty} \frac{e^{100\frac{\ln(n)}{n}}}{10}=\frac{1}{10}$$ thus ...

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