Closed form for infinite sum?

Question

Consider:

$$ 1+1/2^2+2/3^2+1/4^2+2/5^2+1/6^2+...$$

Does this sum have a closed form?

If all the numerators are $1$ then it does have a closed form.

Answer 1

You have

$$1+\dfrac{2-1}{2^2}+\dfrac2{3^2}+\dfrac{2-1}{4^2}+\dfrac2{5^2}+\cdots$$

or

$$2\left(1+\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+\dfrac1{5^2}+\cdots\right)-1-\left(\dfrac1{2^2}+\dfrac1{4^2}+\dfrac1{6^2}+\dfrac1{8^2}+\cdots\right)$$

which is

$$2\left(1+\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+\dfrac1{5^2}+\cdots\right)-1-\frac1{2^2}\left(1+\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+\cdots\right).$$

The rest is obvious.

Answer 2

Yes it does.

Hint: $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$

Hint: $\sum \frac{1}{(2n)^2} = \frac{\pi^2}{24}$

Answer 3

We have $$\frac{\pi^2}{6}=\sum_{n=1}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty \frac{1}{(2n-1)^2}+\sum_{n=1}^\infty\frac{1}{(2n)^2}$$

But $\sum_{n=1}^\infty\frac{1}{(2n)^2}=\frac{1}{4}\sum_{n=1}^\infty\frac{1}{n^2} $ so $$\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$

Note that in general for two real numbers $a,b>0$ we have

$$\sum_{n=1}^{\infty}\frac{1}{(an+b)^2}=\int_0^\infty \frac{te^{-at}}{1-e^{-bt}}dt$$ Then answer is $\frac{\pi^2}{8} + \frac{\pi^2}{6} -1$ .