Calculate the sum $$\sum_{n=1}^{\infty}\Big((2n-1)\Big(\sum_{k=n}^{\infty}\frac{1}{k^2}\Big)-2\Big) $$ The only thing I can think of is that $\sum_{n=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$ However, I can't go any further.
Calculate the sum $\sum_{n=1}^{\infty}\Big((2n-1)\Big(\sum_{k=n}^{\infty}\frac{1}{k^2}\Big)-2\Big) $
Asked
Active
Viewed 75 times
1
-
Where does this task come from? – PinkyWay Mar 27 '20 at 16:46
-
@ms._VerkhovtsevaKatya From a contest – VIVID Mar 27 '20 at 16:51
-
1You should include the link and write down what you have learnes so far. – PinkyWay Mar 27 '20 at 16:52
-
@ms._VerkhovtsevaKatya My source is my friend :) He sent me this for my help, but I couldn't deal with this one – VIVID Mar 27 '20 at 16:55
-
Interesting link for the Basel problem :https://math.stackexchange.com/q/8337/721644 – PinkyWay Mar 27 '20 at 17:01
-
@metamorphy My friend says that that contest is called "The 3rd Open Mathematical Olympiad for University Students" and it's not that famous. – VIVID Mar 27 '20 at 17:08
-
1If it is active, this question will be deleted (most probably). – metamorphy Mar 27 '20 at 17:10
-
@metamorphy No, it is not. It has ended approx. 3-4 days ago – VIVID Mar 27 '20 at 17:11
-
Hint: $$\sum_{n=1}^N (2n-1)\sum_{k=n}^\infty \frac1{k^2} = \sum_{k=1}^\infty \frac1{k^2} \sum_{n=1}^{\min{k,N}} (2n-1) = \sum_{k=1}^N 1 + \sum_{k=N+1}^\infty \frac{N^2}{k^2}.$$ – Greg Martin Mar 27 '20 at 17:12
-
1@metamorphy It would seem the olympiad is over but the solutions are not posted yet. – Simply Beautiful Art Mar 27 '20 at 17:21
1 Answers
2
For $ n\in\mathbb{N} $, define the sequence $ \left(u_{n,k}\right)_{k} $ as follows : $ u_{n,k}=\left\lbrace\begin{aligned}\frac{2n-1}{k^{2}\left(4k^{2}-1\right)},\ \ \ \textrm{If }k\geq n\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{If }k<n\end{aligned}\right. $
Since $ \sum\limits_{n=1}^{+\infty}{u_{n,k}}=\frac{1}{k^{2}\left(4k^{2}-1\right)}\sum\limits_{n=1}^{k}{\left(2n-1\right)}=\frac{1}{4k^{2}-1} $, and $ \sum\limits_{k=1}^{+\infty}{\frac{1}{4k^{2}-1}}=\frac{1}{2}\sum\limits_{k=1}^{+\infty}{\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)}=\frac{1}{2} $, the family $ \left(u_{n,k}\right)_{\left(n,k\right)\in\left(\mathbb{N}^{*}\right)^{2}} $ is summable, and Fubini's theorem allows us to write the following : \begin{aligned} \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\sum_{k=1}^{+\infty}{\sum_{n=1}^{+\infty}}{u_{n,k}}\\ &=\sum_{k=1}^{+\infty}{\frac{1}{4k^{2}-1}}\\ \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\frac{1}{2} \end{aligned}
But since \begin{aligned} \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\sum_{n=1}^{+\infty}{\left(2n-1\right)\sum_{k=n}^{+\infty}{\frac{1}{k^{2}\left(4k^{2}-1\right)}}}\\&=\sum_{n=1}^{+\infty}{\left(2n-1\right)\sum_{k=n}^{+\infty}{\left(\frac{4}{4k^{2}-1}-\frac{1}{k^{2}}\right)}}\\ &=\sum_{n=1}^{+\infty}{\left[\left(2n-1\right)\left(2\sum_{k=n}^{+\infty}{\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)}\right)-\left(2n-1\right)\sum_{k=n}^{+\infty}{\frac{1}{k^{2}}}\right]}\\ \sum_{n=1}^{+\infty}{\sum_{k=1}^{+\infty}{u_{n,k}}}&=\sum_{n=1}^{+\infty}{\left(2-\left(2n-1\right)\sum_{k=n}^{+\infty}{\frac{1}{k^{2}}}\right)} \end{aligned}
We get, $$ \sum_{n=1}^{+\infty}{\left(\left(2n-1\right)\left(\sum_{k=n}^{+\infty}{\frac{1}{k^{2}}}\right)-2\right)}=-\frac{1}{2} $$
CHAMSI
- 8,333