On the Convergence/Divergence of Two Different Series

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.

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

For the first one use limit comparison theorem.

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

(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.

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 ...