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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?

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    $\begingroup$ You made a mistake. $\sum_{p\ge 0, q \ge 0} c_{p+q} = \sum_{m \ge 0} \sum_{p = 0}^m c_{p+(m-p)} = \sum_{m \ge 0} (m+1) c_m$ $\endgroup$
    – reuns
    Commented Oct 25, 2017 at 4:13
  • $\begingroup$ What is the source of this problem? $\endgroup$
    – Carl Schildkraut
    Commented Oct 25, 2017 at 4:22

2 Answers 2

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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.

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  • $\begingroup$ Nice but how did you convert double summation to single summation?because $$S=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty} 2^{-p-q} \int_{0}^{1}t^{p+q}dt$$ $\endgroup$
    – Ekaveera Gouribhatla
    Commented Oct 25, 2017 at 6:10
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    $\begingroup$ @EkaveeraKumarSharma It is double summation: $\sum_{m,n=0}^\infty$ is the same as $\sum_{m=0}^\infty\sum_{n=0}^\infty$. $\endgroup$
    – Angina Seng
    Commented Oct 25, 2017 at 6:54
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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:

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

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    $\begingroup$ You need to first say the series is absolutely convergent, so we can sum it in any order. Then do it diagonally to get a geometric series. $\endgroup$
    – Zhuoran He
    Commented Oct 25, 2017 at 4:27
  • $\begingroup$ @ZhuoranHe This change of summation is essentially that process, is it not? And I'll clarify that the series is absolutely convergent in my answer; thanks for the correction. $\endgroup$
    – Carl Schildkraut
    Commented Oct 25, 2017 at 4:29

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