Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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