# Trying to prove $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$

It has been more than 7 days I have been trying to prove this following result using Harmonic Numbers Let me add this

Proving $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$ doesn't answer my question as I'm looking for the proof-based using Harmonic Numbers. If possible, I would like to avoid integration for the series so that I at least could get the intuitional background.

• My attempts: I could prove

The LHS: $$\left(1-\frac13+\frac15-\frac17+\cdots\right)^2= \lim_{x\to 1}(\tan^{-1}x)^2$$ which because of the this square of the series I was looking for a non-squared series which later

$$\tan^{-1}\frac1x = \sum_{k\ge0}\frac{(-1)^k}{(2k+1)x^{2k}}$$

Which later for $$(\tan^{-1}x)^2 =(\sum_{k\ge0}\frac{(-1)^k}{(2k+1)x^{2k}} )^2$$ Or $$(\tan^{-1}x)^2 = \sum_{q\ge0}\sum_{p\ge0} \frac {(-1)^{p+q}}{(2p+1)(2q+1)}x^{2{(p+q+1)}}$$

In the end I could find this beautiful formula(For me at least) $$\color{blue}{(\tan^{-1}(x))^2 = \sum_{k =1}^{\infty}\frac {(-1)^{k-1}}{k}h_kx^{2k} }$$

$$\text{ Where } h_k = \sum_{i =1}^{k}\frac 1{2i-1} = H_{2k} - \frac12H_k$$

Which later using $$\sum_{k =1}^x a_kb_k = S_xb_x - \sum_{k =1}^{x-1}S_k(b_{k+1} - b_k)$$

$$\sum_{k =1}^\infty\frac{(-1)^{k-1}(H_{2k} - \frac12H_{k})}{k} = \lim_{x\to\infty}\left(H'_x(H_{2x} -\frac12H_x) - \sum_{k =1}^{x -1}\frac {H'_k}{2k+1}\right)$$

I tried to use a few other identities so that I at least could end up to $$\zeta(2), Li_n(x)$$ but I'm stuck and don't know how to go forward

Also,

$$f(x) = \sum_{k =1}^{\infty}\frac {h_kx^{2k-1}}{2k-1} \text { , } |x|< 1$$ $$f\left(\frac x{2-x}\right) = \frac18\log^2(1-x) + \frac12Li_2(x)$$ Where, $$Li_n(1) = \zeta(n)$$

• I do not want to use facts like $$\frac π4,\frac{π^2}{6}$$ I'm looking for the complete proof that bridges square of alternate odd harmonic number series with a sum of reciprocals of squared natural numbers.

• Why did I add this problem to MSE? I'm really not able to find time for solving maths and can't live stable without solving it at the same time. Even If I'll have time most of all my time goes into finding things that are already(invented) as I lack mathematical background most of all of the above formulas are from Ramanujan's lost notebook and most of all my time goes in shuffling pages... I Hope MSE can help me live obsessed-free life...haha

• I don't want to be get upvoted but please try to add your valuable suggestions before downvoting or else if it's that easy to prove using Harmonic Numbers then please don't forget you are always welcome to add your answer I'll bounty all my reputations if answered < 7days else: all upvotes of this question: Commented Nov 18, 2021 at 7:37
• Does this answers your question ? Commented Nov 18, 2021 at 12:50
• @Tryhard It would but isn't devoted to Harmonic# also uses substitution over integrals which (sorry)lacks the fragrant sense of intuition(at least for me) Commented Nov 18, 2021 at 12:57
• I think you forgot to add the tags " harmonic-number" and " alternative-prood" . Commented Nov 18, 2021 at 13:32
• So basically you want to prove ( by series manipulation) $\sum_{n=1}^\infty (-1)^{n-1} \frac{2H_{2n}-H_n}{n}=\frac34\sum_{n=1}^\infty \frac1{n^2}$ Commented Nov 18, 2021 at 15:11

Use Newton-Gregory series $$\tan^{-1}1=1-1/3+1/5-1/7+...=\frac{\pi}{4}$$ and a series due to Fourier theorem $$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$
• Thanks, but refer it is to be proved disregarding the fact that $\frac {\pi}{4}$ Commented Nov 18, 2021 at 9:32
• The Gregory series is elementary but finding $\sum_n 1/n^2$ is deeper, which is why the OP did not succeed. Commented Nov 18, 2021 at 9:35
• @DanielWainfleet Exactly Daniel, However, I know almost all possible ways to find $\sum\frac1{n^2} = \frac {\pi^2}{6}$ but sorry I don't want to use these facts. Commented Nov 18, 2021 at 9:39