Convergence of Series
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Convergence of Series
Examine whether the following series converges or not: \( \displaystyle \sum_{n=2}^{\infty}\left ( \frac{1}{\left(n\ln n\right)^2}-n \right ) \).
Imagination is much more important than knowledge.
Re: Convergence of Series
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!
\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!
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 0 guests