<s>[INST] Correctness of proof for the convergence of a series Does the following series converge?
$\sum_{n=1}^\infty \frac{5^n + (-1)^n}{2^n3^n}$
What I've done so far.
$0\leq \sum_{n=1}^\infty \frac{5^n + (-1)^n}{2^n3^n} \leq \sum_{n=1}^\infty \frac{5^n + 1^n}{6^n} = \sum_{n=1}^\infty \left(\frac{5}{6}\right)^n + \sum_{n=1}^\infty \left(\frac{1}{6}\right)^n$
We know that last two series both converge, since they are an instance of a geometric series with both $|q|<1$.
Now from this, am I allowed to conclude that the original series converges? What stops it from alternating between 0 and the sum of the values of the geometric series?
 [/INST] Notice, we have $$\sum_{n=1}^{\infty}\frac{5^n+(-1)^n}{2^n\cdot 3^n}$$ $$=\sum_{n=1}^{\infty}\frac{5^n+(-1)^n}{6^n}$$
$$=\sum_{n=1}^{\infty} \frac{5^n}{6^n}+\sum_{n=1}^{\infty}\frac{(-1)^n}{6^n}$$ 
$$=\sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n+\sum_{n=1}^{\infty}\left(\frac{-1}{6}\right)^n$$ 
$$=\left(\frac{5}{6}+\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)^3+\ldots\right)+\left(\frac{-1}{6}+\left(\frac{-1}{6}\right)^2+\left(\frac{-1}{6}\right)^3+\ldots\right)$$ $$=\left(\frac{\frac{5}{6}}{1-\frac{5}{6}}\right)+\left(\frac{-\frac{1}{6}}{1-\left(-\frac{1}{6}\right)}\right)$$ 
$$=\left(\frac{\frac{5}{6}}{\frac{1}{6}}\right)+\left(\frac{-\frac{1}{6}}{\frac{7}{6}}\right)$$ 
$$=5-\frac{1}{7}=\frac{35-1}{7}$$ $$=\frac{34}{7}$$ Above is a finite value. Hence the series converges.  
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