Squaring a Leibnitz-like series

The series

$$\frac{\pi^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\tag1$$

can serve as either a corollary or an antecedent to the familiar Basel problem series

$$\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$$

Now suppose we take the square root of ${\pi^2/8}$ getting $\pi/(2\sqrt2)$. This has its own Leibnitz-like series

$$\dfrac{\pi}{2\sqrt2}=\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\cdots\tag2$$

where the sign reverses when the denominator crosses a multiple of $4$. This latter series is provable in many ways, for instance with a Fourier series.

Since the left side of Eq. 1 is the square of the left side of Eq. 2, it is tempting to suppose that somehow Eq. 1 could be derived by squaring Eq. 2. A brute force approach generates the right squared terms, but how do the cross-product terms cancel out to get the correct form for Eq. 1?

I have done some work on the problem, identifying a method that appears to derive Eq. (2) from Eq. (1). The method is based on the "long division" method of square root extraction.

Assume that the square root of

$$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots$$

is to be rendered in the form

$$s_0\dfrac{1}{1}+s_1\dfrac{1}{3}+s_2\dfrac{1}{5}+s_3\dfrac{1}{7}+\cdots$$

where each $s_k$ term is $\pm1$ and the signs are chosen to give back the original sum of squares with no cross-product terms.

As with the usual version of the square root extraction algorithm, we begin with the square root of the first term of the series, thus $+1$ making the $s_0$ sign positive. We multiply this by $2$ to get$2$, which we use later.

Now compare the next term in the radicand, $+1/9$, with $-1/2$ times the next term whose sign is to be determined, $1/3$. The comparison figure is meant to compensate approximately for the rest of the infinite series in the radicand; that is for each odd $n$ the sum of terms following $1/n^2$ is approximated as $1/(2n)$ and our comparison figure of $-1/(2n)$ will compensate for that. Clearly $+1/9$ is greater, indicating that the exact square root will be greater than our current partial sum ($1$). Like a one-dimensional game of "hot and cold", we therefore select $s_1=+1$, making our second partial sum $1+(1/3)$ and its doubled value, $2+(2/3)=8/3$, to be used in the next round.

Now add the $+1/3$ term we just rendered to the doubled value of the pervious sum, giving $2+(1/3)=7/3$, then multiply by $+1/3$ to get $7/9$. Subtract this from $+1/9$, which was our previous leading term, and combine with the next term $1/25$ from the radicand to get $-49/75$ as our next leading term. This is now less than $-1/2$ times the $1/5$ term we are signing next, so we assume our square root must be less than $1+(1/3)$ and thus render $s_2=-1$, giving $1+(1/3)-(1/5)$ as our third partial sum and twice that equaling $(8/3)-(2/5)=34/15$.

Adding $-1/5$ to $34/15$ gives $31/15$ and multiplying the latter by the $-1/5$ term we introduced gives $-31/75$, which when subtracted from $-49/75$ yields $-6/25$. Combining this with $1/49$ gives $-269/1225$ as our next leading term. This is still less than $(-1/2)×(1/7)=-1/14$, so we will assign $s_3=-1$ for the $1/7$ term in our square root.

The sign pattern thus far ($+(1/1)+(1/3)-(1/5)-(1/7)$) matches that for the series in Eq. (2), and when I set up the procedure on a MS Excel spreadsheet the correct sign pattern is found to hold for at least 101 terms. This approach can be converted to a proof of the claim if we can prove that:

• The algorithm described above will continue to give the sign cycle $++--$.

• Whenever we truncate the square root series at a positive term it's square will be greater than the given radicand, but the opposite is true when we truncate at a negative term.

This problem is more complex than just squaring $\frac{\pi}{2\sqrt{2}}$

We begin with :

$$ \frac{\pi}{2\sqrt{2}} = \sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac{1}{4n-3}+\frac{1}{4n-1}\right)$$

then $$ \frac{\pi^2}{8} = \left(\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac{1}{4n-3}+\frac{1}{4n-1}\right)\right)^2 = \left(2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{4n-3} \,- \frac{\ln{(1+\sqrt{2})}}{\sqrt{2}}\right)^2 $$

and we end up with

$$ \frac{\pi^2}{8} = -\frac{\ln(1+\sqrt{2})^2}{2} - \frac{\pi \, \ln(1+\sqrt{2})}{2} + 4\left(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{4n-3}\right)^2 $$

Squaring the sum as given in How to Evaluate $ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} $

$$ \frac{\pi^2}{8} = -\frac{\ln(1+\sqrt{2})^2}{2} - \frac{\pi \, \ln(1+\sqrt{2})}{2} + 4\sum_{n=1}^{\infty} \frac{1}{(4n-3)^2}+2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k}\frac{1}{4n-3} \tag{1} $$

Now , using similar techniques we have :

$$ \frac{\pi^2}{8} = -\frac{\ln(1+\sqrt{2})^2}{2} + \frac{\pi \, \ln(1+\sqrt{2})}{2} + 4\sum_{n=1}^{\infty} \frac{1}{(4n-1)^2}+2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k}\frac{1}{4n-1} \tag{2}$$

Add $(1)\,+\,(2)$

$$ \frac{\pi^2}{4} = -\ln(1+\sqrt{2})^2 +4\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} + 2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k}\frac{1}{4n-3}+\frac{1}{4n-1}$$

We can split the double sum as

$$ 2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k}\frac{1}{4n-3}+\frac{1}{4n-1} = {2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k}\frac{1}{2n-1}}+ 2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k} \frac{1}{2n+2k-1} $$

I will not show proof of this but :

$$ {2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k}\frac{1}{2k-1}} = -\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} $$

$$ 2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{k} \frac{1}{2n+2k-1} = -\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} + \ln(1+\sqrt{2})^2 $$

We conclude with

$$ \frac{\pi^2}{4} = -\ln(1+\sqrt{2})^2+4\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} - \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} - \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} + \ln(1+\sqrt{2})^2 $$

therefore

$$ \frac{\pi^2}{8} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} $$

NOTE

Using the technique as seen in the link you may create a multitude of infinite series by multiplying 2 convergent infinite series.

If you are looking to square a sum in order to evaluate $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} $$

then you should look at squaring

$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)} = -\frac{\pi}{4} $$

If typos or errors kindly edit or let me know. Thank you.