Find the value of $\sum^{\infty}_{n=0} \sum^{n}_{k=0} \frac{2^k}{2^n (k+1)(k+2)}.$

Question

Find the value of $$ \sum^{\infty}_{n=0} \sum^{n}_{k=0} \frac{2^k}{2^n (k+1)(k+2)}. $$ I'm guessing this has something to do with the cauchy product so my first thought was to dissect this series into two seperate ones? I've gotten to the point where i have $$ \sum^{\infty}_{n=0} \sum^{n}_{k=0} a_{n-k} b_{k} = \sum^{\infty}_{n=0} \sum^{n}_{k=0} \frac{2^k}{2^n (k+1)(k+2)} $$ where $\sum^{\infty}_{n=0} a_n $ and $\sum^{\infty}_{n=0} b_n$ would be those smaller series within.. now idk how they would be defined as though. (I can also kind of sense that terms will end up cancelling each other out because of the (k+1)(k+2) but I haven't gotten to the point where I can use that yet (or so I think).)

Answer 1

Consider $a_n= \frac{1}{2^n}$ and $b_n= \frac{1}{(n+1)(n+2)}.$

Then we have

$$\sum^{\infty}_{n=0} \sum^{n}_{k=0} a_{n-k} b_{k} = \sum^{\infty}_{n=0} \sum^{n}_{k=0} \frac{2^k}{2^n (k+1)(k+2)}.$$

Hence

$$\sum^{\infty}_{n=0} \sum^{n}_{k=0} \frac{2^k}{2^n (k+1)(k+2)}= (\sum^{\infty}_{n=0}a_n)(\sum^{\infty}_{n=0}b_n) $$

Your turn !