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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.

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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.

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    $\begingroup$ Thanks! This is what I needed. $\endgroup$
    – KaRJ XEN
    Oct 12, 2015 at 23:39
  • $\begingroup$ Glad that I could help :) it is good manners, however, to mark answers as accepted. $\endgroup$ Oct 12, 2015 at 23:42
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    $\begingroup$ Sorry, I forgot to do that before but I just did! Thanks again. $\endgroup$
    – KaRJ XEN
    Oct 13, 2015 at 10:45
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$$ X = \sum_{k=0}^\infty \left(\frac{z}{5}\right)^k \sum_{k=0}^\infty \left(\frac{2z}{5}\right)^k$$ 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, $$|z| < \frac{5}{2} \,\ \text{holds true.}$$

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

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