Cauchy Product of two simple geometric series

Question

I want to calculate the cauchy product of two simple geometric series:

$$\left(\sum_{n=0}^\infty (1/2)^n \right)\left(\sum_{n=0}^\infty (1/5)^n \right) = 5/2 $$ Wolfram Alpha result

According to Wolfram Alpha and finding not the right answer, my calculation is supposed to be wrong: $$\left(\sum_{n=0}^\infty (1/2)^n \right)\left(\sum_{n=0}^\infty (1/5)^n \right) = \sum_{n=0}^\infty \sum_{k=0}^n 1/(2^n5^{n-k}) $$ Wolfram Alpha

Can somebody give me a hint, why this is the wrong equatation?

Answer 1

In general ...

$$\left(\sum_{n=0}^\infty a_n \right)\left(\sum_{n=0}^\infty b_n \right) = \\\sum_{n=0}^\infty \sum_{k=0}^n a_nb_{n-k}+ \sum_{n=0}^\infty \sum_{k=1}^{n} a_{n-k}b_{n}$$

Where the first sum adds terms $a_ib_j$ with $i\ge j$ and the second adds terms $i<j$

A more symmetrical form ( and the one is used to do your calculation ) is ...

$$\left(\sum_{n=0}^\infty a_n \right)\left(\sum_{n=0}^\infty b_n \right) = \\\sum_{n=0}^\infty \sum_{k=0}^n a_nb_{n-k}+ \sum_{n=0}^\infty \sum_{k=0}^{n} a_{n-k}b_{n} - \sum_{n=0}^\infty a_n b_n \\= \sum_{n=0}^\infty \sum_{k=0}^n ( a_nb_{n-k}+a_{n-k}b_n)- \sum_{n=0}^\infty a_n b_n$$

where the last term is correcting for double counting the $i=j$ case

in your case you can verify that ...

$$\left(\sum_{n=0}^\infty (1/2)^n \right)\left(\sum_{n=0}^\infty (1/5)^n \right) = \sum_{n=0}^\infty \sum_{k=0}^n 1/(2^n5^{n-k}) + \sum_{n=0}^\infty \sum_{k=0}^n 1/(2^{n-k}5^{n}) - \sum_{n=0}^\infty 1/(2^n5^{n}) =\frac 52$$

Answer 2

In general... $$\left(\sum_{k=0}^\infty a_k\right)\left(\sum_{j=0}^\infty b_j\right) =\sum_{n=0}^\infty\left(\sum_{k=0}^na_kb_{n-k}\right)$$ (for more details about convergence conditions, see Cauchy product).

In the case of two convergent geometric series like yours, with $\alpha,\beta,0$ distinct (in your case: $\alpha=\frac12,\beta=\frac15$), you can check that... $$\begin{align}\left(\sum_{k=0}^\infty\alpha^k\right)\left(\sum_{j=0}^\infty\beta^j\right)&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\alpha^k\beta^{n-k}\right)\\&=\sum_{n=0}^\infty\beta^n\left(\sum_{k=0}^n(\alpha/\beta)^k\right)\\&=\sum_{n=0}^\infty\beta^n\frac{(\alpha/\beta)^{n+1}-1}{(\alpha/\beta)-1}\\&=\frac1{\alpha-\beta}\left(\alpha\sum_{n=0}^\infty\alpha^n-\beta\sum_{n=0}^\infty\beta^n\right)\\&=\frac1{\alpha-\beta}\left(\frac\alpha{1-\alpha}-\frac\beta{1-\beta}\right)\\&=\frac1{(1-\alpha)(1-\beta)}.\end{align}$$ Note the RHS in the first equality: there is an $\alpha^k$, not an $\alpha^n$ like in your question and in WW1's answer.

For the case $\alpha=\beta$, see Cauchy product of two geometric series.