# Product of two sums over the same interval

I have some terms of an expression as sums but I would like to simplify the solution to an easier and less complicated one. What I have is $$X = \sum_{k=0}^\infty \left(\frac{z}{5}\right)^k \sum_{r=0}^\infty \left(\frac{2z}{5}\right)^r$$ Can this be reduced to a single sum (It should be an infinite sum and not just an expression)? Any help is greatly appreciated. Thanks.

First of all, you should be aware of the radius of convergence for the sum (the sum is convergent for $|z|<\frac{5}{2}$). Another thing to point out is that it's wise to use different indices for those two sums to avoid confusion (e.g. sum over $k$ and $j$).

That being said, you may write the product as $$X = \sum_{k=0}^\infty \left(\frac{z}{5}\right)^k \sum_{j=0}^\infty \left(\frac{2z}{5}\right)^j$$ and proceed with the formula for the sum of the infinite geometric series $$\sum_{k=0}^\infty aq^k=\frac{a}{1-q},$$ as the answer by Aniket elaborates.

If you want to reduce it to a single sum, you may take the finished expression and expand it into an infinite sum (Taylor/McLaurin/Laurent series, depending on what you want to achieve).

What you can also do is write down the first expression using the polynomial multiplication formula: $$A=\sum_{k=0}^\infty \alpha_k x^k, \space B=\sum_{j=0}^\infty \beta_j x^j$$ $$\Rightarrow C=AB=\sum_{n=0}^\infty \left(\sum_{k+j=n}\alpha_k \beta_j\right)x^n$$ Note that here you have to choose the same $x$ for sum $A$ and $B$ for the formula to be correct (e.g. $x=\frac{z}{5}$). The coefficients are then $\alpha_k=1$ and $\beta_j=2^j$ so the final expression can be simplified to $$X=\sum_{n=0}^\infty \left(\frac{z}{5}\right)^n \sum_{j=0}^{n}2^j=\sum_{n=0}^\infty\left[ \left(\frac{z}{5}\right)^n\left(2^{n+1}-1\right)\right]$$ You can easily check that this sum yields the same closed form as Aniket's result.

• Thanks! This is what I needed. Oct 12, 2015 at 23:39
• Glad that I could help :) it is good manners, however, to mark answers as accepted. Oct 12, 2015 at 23:42
• Sorry, I forgot to do that before but I just did! Thanks again. Oct 13, 2015 at 10:45

$$X = \sum_{k=0}^\infty \left(\frac{z}{5}\right)^k \sum_{k=0}^\infty \left(\frac{2z}{5}\right)^k$$ or, $$X = \left(\frac{1}{1 - \frac{z}{5}} \right) \cdot\left (\frac{1}{1 - \frac{2z}{5}} \right)$$ or, $$X = \frac{25}{(5-z)(5-2z)}$$ Provided $z \not = 5, \frac{5}{2}$ and the individual sums are themselves convergent i.e. $|\frac{z}{5}|\ < 1$ and $|\frac{2z}{5}|\ < 1$ or, in other words, $$|z| < \frac{5}{2} \,\ \text{holds true.}$$

• $X_k$ should be renamed $X$, or anything not depending on $k$.
– Did
Oct 12, 2015 at 19:48
• @Did Thank you so much for pointing out my incorrect answer! It has since been deleted at my request. I really appreciate your help. Oct 12, 2015 at 19:57