Sum $\sum_{x=1}^n\sum _{y=1}^{x-1}\frac{1/2^x*1/2^y}{1/2^x+1/2^y}$

Question

Is there a way to calculate following summation

$\sum_{x=1}^n\sum _{y=1}^{x-1}\frac{1/2^x*1/2^y}{1/2^x+1/2^y}$

Can it be reduced to something simple?

Answer 1

We can pretty simply reduce the limit as $n\to\infty$ to a single sum: $$ \begin{align} \sum_{k=1}^\infty\sum_{j=1}^{k-1}\frac1{2^j+2^k} &=\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac1{2^j+2^k}\tag{1}\\ &=\sum_{j=1}^\infty\frac1{2^j}\sum_{k=j+1}^\infty\frac1{2^{k-j}+1}\tag{2}\\ &=\sum_{j=1}^\infty\frac1{2^j}\sum_{k=1}^\infty\frac1{2^k+1}\tag{3}\\ &=\sum_{k=1}^\infty\frac1{2^k+1}\tag{4} \end{align} $$ Explanation:
$(1)$: change order of summation
$(2)$: factor $2^{-j}$ out of the inner sum
$(3)$: substitute $k\mapsto k+j$
$(4)$: sum in $j$

I don't see an easy way to simplify the finite sum.

Answer 2

There's probably no closed form. However, the double sum converges quite rapidly. Denoting your sum by $T_n$, it is more natural to consider instead the sum $$ S_n = \sum_{x,y=1}^n \frac{1}{2^x+2^y}. $$ You can check that $$ T_n = \frac{S_n - \sum_{x=1}^n \frac{1}{2^{x+1}}}{2} = \frac{S_n - \frac{1}{2} + \frac{1}{2^{n+1}}}{2}. $$ We can estimate the infinite sum $$ \begin{align*} S &= \sum_{x,y=1}^\infty \frac{1}{2^x+2^y} \\ &\leq \sum_{x,y=1}^\infty \frac{1}{2^{\max(x,y)}} \\ &\leq \sum_{m=1}^\infty \frac{2m+1}{2^m} = 5. \end{align*} $$ The actual value is $S \approx 2.02899956069689$.

The infinite limit of your sum is $T = (S-\tfrac{1}{2})/2 \approx 0.764499780348444$.

How fast is the convergence? $$ \begin{align*} S - S_n &\leq 2\sum_{x=n+1}^\infty \sum_{y=1}^\infty \frac{1}{2^x+2^y} \\ &\leq 2\sum_{x=n+1}^\infty \sum_{y=1}^\infty \frac{1}{2^{\max(x,y)}} \\ &\leq 2\sum_{m=n+1}^\infty \frac{2m+1}{2^m} \\ &= O\left(\frac{n}{2^n}\right). \end{align*} $$ You can similarly obtain a bound from the other direction, and deduce that $S_n = S - \Theta(\frac{n}{2^n})$. Therefore $T_n = T - \Theta(\frac{n}{2^n})$.

We can obtain another expression for $T$ using the identity $$ \frac{1}{2^x+2^y} = \frac{1}{2^x} \sum_{n=0}^\infty (-1)^n \frac{1}{2^{n(x-y)}}. $$ Using this, we get $$ \begin{align*} T &= \sum_{x=1}^\infty \sum_{y=1}^{x-1} \frac{1}{2^x+2^y} \\ &= \sum_{x=1}^\infty \sum_{y=1}^{x-1} \frac{1}{2^x} \sum_{n=0}^{\infty} (-1)^n \frac{1}{2^{n(x-y)}} \\ &= \sum_{n=0}^\infty (-1)^n \sum_{x=1}^\infty \frac{1}{2^x} \sum_{y=1}^{x-1} \frac{1}{2^{n(x-y)}} \\ &= \sum_{x=1}^\infty \frac{x-1}{2^x} + \sum_{n=1}^\infty (-1)^n \sum_{x=1}^\infty \frac{1}{2^{(n+1)x}} \sum_{y=1}^{x-1} 2^{ny} \\ &= 1 + \sum_{n=1}^\infty (-1)^n \sum_{x=1}^\infty \frac{1}{2^{(n+1)x}} \frac{2^{nx}-2^n}{2^n-1} \\ &= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} \left[\sum_{x=1}^\infty \frac{1}{2^x} - 2^n \sum_{x=1}^\infty \frac{1}{2^{(n+1)x}} \right] \\ &= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} \left[1 - \frac{1}{2-2^{-n}}\right] \\ &= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2^{n+1}-1} \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}-1}. \end{align*} $$ We can obtain another expression by opening $1/(2^{n+1}-1)$ into a series: $$ \begin{align*} T &= \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}-1} \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}} \sum_{m=0}^\infty \frac{1}{2^{(n+1)m}} \\ &= \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(-1)^n}{2^{(m+1)(n+1)}} \\ &= \sum_{m=0}^\infty \frac{1}{2^{m+1}+1}. \end{align*} $$ Summarizing, $$ T = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}-1} = \sum_{m=0}^\infty \frac{1}{2^{m+1}+1}. $$