<s>[INST] Given $\sum_{n=0}^\infty \frac{1}{2^n}$ and $\sum_{n=0}^\infty \frac{1}{4^n}$, what is the Cauchy product? Given $\sum_{n=0}^\infty \frac{1}{2^n}$ and $\sum_{n=0}^\infty \frac{1}{4^n}$, what is the Cauchy product?
The definition of the Cauchy product is for two given series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$: $$c_n = \sum_{j=0}^n a_j b_{n-j}$$.
With the two series I would get $c_n = \sum_{j=0}^n \frac{1}{2^j} \frac{1}{4^{n-j}} = \sum_{j=0}^n \frac{1}{2^{-j}2^n}$ How can I simplify this to a closed form?
 [/INST] I think you mean
$c_n=\sum\limits_{j=0}^n\frac{1}{2^j}\frac{1}{4^{n-j}}=\sum\limits_{j=0}^n\frac{1}{2^j}\frac{1}{2^{2n-2j}}=\sum\limits_{j=0}^n\frac{1}{2^{2n-j}}=\sum\limits_{j=0}^n2^{j-2n}=2^{-2n}\sum\limits_{j=0}^n2^j$
The sum on the right hand side is a geometric series, which has the formula
$\sum\limits_{j=0}^nar^j=\frac{a(1-r^{n+1})}{1-r}$
Plugging in $a=1,r=2$, we have
$c_n=2^{-2n}\sum\limits_{j=0}^n2^j=2^{-2n}(\frac{1-2^{n+1}}{1-2})=\frac{2^{n+1}-1}{2^{2n}}$
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