G285 2021 Fall Problem Set Problem 8

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Problem

If the value of \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c)}}\] can be represented as $\frac{m}{n}$, where $m$ and $n$ are relatively prime. Find $m+n$.

Solution

We begin with a simpler problem \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{1}{4^{(a+b+c)}}\]. Now, suppose $a$ and $b$ are constant. We have a converging geometric series for $c$ with a sum of $\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$. Now, make $b$ everchanging. We have $\frac{1}{4^{b+c})=\left(\frac{4}{3} \right)^2 = \frac{16}{9}$ (Error compiling LaTeX. Unknown error_msg), so the entire sum must be $\frac{64}{27}$.

Now, coming back to the original problem, we split the single sum into $3$: \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c)}}\]