Prove or disprove: The Cauchy-product of series $\sum_{k=0}^{\infty}\frac{1}{2^{k}}$ and $\sum_{k=0}^{\infty}\frac{1}{4^{k}}$ equals...

Question

Prove or disprove: The Cauchy-product of series $\sum_{k=0}^{\infty}\frac{1}{2^{k}}$ and $\sum_{k=0}^{\infty}\frac{1}{4^{k}}$ equals $\sum_{k=0}^{\infty}\left(2^{1-k}-2^{-2k} \right )$

$$\sum_{k=0}^{\infty}\sum_{k}^{l}\frac{1}{2^{l}}\cdot \frac{1}{4^{k-l}}=\sum_{k=0}^{\infty}\sum_{k}^{l}2^{-l}\cdot 4^{-k+l}= \sum_{k=0}^{\infty}\sum_{k}^{l}2^{-l}\cdot 4^{-k}\cdot 4^{l}= \sum_{k=0}^{\infty}2^{l}\cdot 4^{-k}$$

As it looks like, the statement is false.

Did I do everything correctly?

Answer 1

The actual Cauchy product is supposed to be:$$ \sum_{k=0}^\infty\sum_{l=0}^k 2^{-l}4^{l-k}= \sum_{k=0}^\infty\sum_{l=0}^k 2^{-l}2^{l-k} 2^{l-k} =\sum_{k=0}^\infty 2^{-2k}\sum_{l=0}^k 2^l=\sum_{k=0}^\infty 2^{-2k}(2^{k+1}-1) $$

So the statement is true.

Answer 2

Statement is true. $$S=\sum_{k=0}^{\infty}\sum_{t=0}^{k}(\frac{1}{2^{t}}\cdot \frac{1}{4^{k-t}})=\sum_{k=0}^{\infty}\sum_{t=0}^{k}\frac{1}{2^{2k-t}}$$ Since $$\sum_{t=0}^{k}\frac{1}{2^{2k-t}}=\frac{1}{2^k}\frac{1-\frac{1}{2^{k+1}}}{1-\frac{1}{2}}=\frac{2^{k+1}-1}{2^{2k}},$$ we have $$S=\sum_{k=0}^{\infty} \frac{2^{k+1}-1}{2^{2k}}=2\sum_{k=0}^{\infty} \frac{1}{2^{k}}-\sum_{k=0}^{\infty}\frac{1}{2^{2k}}$$ $$S=2\frac{1}{1-\frac{1}{2}}-\frac{1}{1-\frac{1}{4}}=4-\frac{4}{3}=\frac{8}{3}$$