## anonymous one year ago Can you please show me how to solve these and send me the solution on skype? (my skype ID is golemboy1 from Brasov Romania) http://tinypic.com/r/nl1r93/8

1. thomas5267

$\sum_{n=1}^\infty \frac{n}{n^2+1}\geq\sum_{n=1}^\infty \frac{n}{n^2+n}=\sum_{n=1}^\infty \frac{1}{n+1}=\sum_{n=2}^\infty \frac{1}{n}\\$ Harmonic series diverges.

2. thomas5267

$\sum_{n=1}^\infty\frac{n}{n^3+1}\leq\sum_{n=1}^\infty\frac{n}{n^3}=\sum_{n=1}^\infty\frac{1}{n^2}\\ \int_1^\infty \frac{1}{n^2}\,dn=\left[-\frac{1}{n}\right]_1^\infty=1\\ \sum_{n=1}^\infty\frac{n}{n^3+1} \text{ converges.}$

3. thomas5267

\sum_{n=1}^\infty\frac{n^2}{2^n}\\ \begin{align*} &\phantom{{}={}}\lim_{n\to\infty}\frac{(n+1)^2}{2^{n+1}}\frac{2^n}{n^2}\\ &=\lim_{n\to\infty}\frac{n^2+2n+1}{2n^2}\\ &=\frac{1}{2} \end{align*} \\ \sum_{n=1}^\infty\frac{n^2}{2^n}\text{ converges and trivially converges absolutely since all terms are positive.}

4. thomas5267

18a and b are direct calculation so I will leave this for you.

5. thomas5267

$\sum_{n=1}^\infty\frac{n+1}{2n+1}\geq\sum_{n=1}^\infty\frac{n+1}{2n+2}=\frac{1}{2}\sum_{n=1}^\infty1\\ \text{You tell me whether this converges or not.}\\ \sum_{n=1}^\infty\frac{1}{2n}=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n}\\ \text{Harmonic series does not converge. Wikipedia the proof.}\\ \sum_{n=1}^\infty\frac{2^n}{n}\geq\sum_{n=1}^\infty\frac{2^n}{2^n}=\sum_{n=1}^\infty1\\ \text{You tell me whether this converges or not.}$

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