Find the value of $\sum_{p=0}^{\infty}\sum_{q=0}^{\infty} \frac{2^{-p-q}}{1+p+q}$

Question

Find the value of

$$S=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty} \frac{2^{-p-q}}{1+p+q}$$

In the second summation i used change of variable $p+q+1=r$ then we get

$$S=\sum_{p=0}^{\infty}\:\sum_{r=p+1}^{\infty}\frac{2^{1-r}}{r}$$ $\implies$

$$S=2\sum_{p=0}^{\infty}\:\sum_{r=p+1}^{\infty}\frac{1}{r2^r}$$

Any clue here?

Answer 1

An alternative method is to recognise the sum as $$S=\int_0^1\sum_{p,q=0}^\infty 2^{-p-q}t^{p+q}\,dt =\int_0^1\frac{dt}{(1-t/2)^2}$$ etc.

Answer 2

Note: As noted by Zhuoran He, to do any change of summation, we first need to verify that the series is absolutely convergent. This can be done via comparison to the sum of $2^{-p-q}$ over that same range.

Reuns is correct. Under the change of variables $p+q+1=r$, we have $r$ values of $(p,q)$ that yield $p+q+1=r$ (as $p$ ranges from $0$ to $r-1$), so the sum becomes very nice indeed:

$$\sum_{r=1}^{\infty} r\left(\frac{2^{1-r}}{r}\right).$$