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 Post subject: Convergence of SeriesPosted: Sat Jul 09, 2016 8:15 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Examine whether the following series converges or not: $\displaystyle \sum_{n=2}^{\infty}\left ( \frac{1}{\left(n\ln n\right)^2}-n \right )$.

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 Post subject: Re: Convergence of SeriesPosted: Fri May 19, 2017 7:01 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
The series diverges. To see that we'll work with partial sums:

\begin{align*}
\sum_{n=2}^{\infty} \left [ \frac{1}{n^2 \log^2 n} - n \right ] &= \lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \left [ \frac{1}{n^2 \log^2 n} - n \right ] \\
&= \sum_{n=2}^{\infty} \frac{1}{n^2 \log^2 n } - \lim_{N \rightarrow +\infty} \sum_{n=2}^{N} n \\
&=\sum_{n=2}^{\infty} \frac{1}{n^2 \log^2 n } - \lim_{N \rightarrow +\infty} \frac{N\left ( N+1\right )}{2} +1 \\
&\rightarrow -\infty
\end{align*}

It is known that the series $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^2 \log^2 n }$ converges.

P.S: A possible interesting question is to examine the convergence of the series

$$\mathcal{S} = \sum_{n=2}^{\infty} \left [ \frac{1}{n \log n} - n \right ]$$

I have not worked it out but I rush to say that my calculator says it diverges!

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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