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

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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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[ \sum_{n=1}^\infty \frac{n}{n^2+1}\geq\sum_{n=1}^\infty \frac{n}{n^2+n}=\sum_{n=1}^\infty \frac{1}{n+1}=\sum_{n=2}^\infty \frac{1}{n}\\ \] Harmonic series diverges.
\[ \sum_{n=1}^\infty\frac{n}{n^3+1}\leq\sum_{n=1}^\infty\frac{n}{n^3}=\sum_{n=1}^\infty\frac{1}{n^2}\\ \int_1^\infty \frac{1}{n^2}\,dn=\left[-\frac{1}{n}\right]_1^\infty=1\\ \sum_{n=1}^\infty\frac{n}{n^3+1} \text{ converges.} \]
\[ \sum_{n=1}^\infty\frac{n^2}{2^n}\\ \begin{align*} &\phantom{{}={}}\lim_{n\to\infty}\frac{(n+1)^2}{2^{n+1}}\frac{2^n}{n^2}\\ &=\lim_{n\to\infty}\frac{n^2+2n+1}{2n^2}\\ &=\frac{1}{2} \end{align*} \\ \sum_{n=1}^\infty\frac{n^2}{2^n}\text{ converges and trivially converges absolutely since all terms are positive.} \]

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18a and b are direct calculation so I will leave this for you.
\[ \sum_{n=1}^\infty\frac{n+1}{2n+1}\geq\sum_{n=1}^\infty\frac{n+1}{2n+2}=\frac{1}{2}\sum_{n=1}^\infty1\\ \text{You tell me whether this converges or not.}\\ \sum_{n=1}^\infty\frac{1}{2n}=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n}\\ \text{Harmonic series does not converge. Wikipedia the proof.}\\ \sum_{n=1}^\infty\frac{2^n}{n}\geq\sum_{n=1}^\infty\frac{2^n}{2^n}=\sum_{n=1}^\infty1\\ \text{You tell me whether this converges or not.} \]

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