<s>[INST] Infinite Geometric Series and its Value Calculate the value of  $\sum_1^\infty (\frac{1}{3})^{2n}$.
I can clearly see that after listing a few partial sums this seems to tend toward $\frac{1}{8}$ and each partial sum is greater than the last.
$\sum_1^2 (\frac{1}{3})^{2n} = \frac{10}{81}$
$\sum_1^3 (\frac{1}{3})^{2n} = \frac{91}{729}$
$\sum_1^4 (\frac{1}{3})^{2n} = \frac{820}{6561}$
$ \frac{10}{81} \lt \frac{91}{729} \lt \frac{820}{6561} \lt {...}\lt \frac{1}{8} $
My question is how do I show that the answer is indeed $\frac{1}{8}$? I considered using a theorem in my book that says that $\sum_1^\infty x_n=\lim(s_k) = sup\{s_k :k \in N\}$ where $s_k$ are partial sums. The issue I am coming into is showing that $\frac{1}{8}$ is indeed the supremum. Any ideas would be appreciated. I also don't have to use this theorem, it was just a thought.
 [/INST] The formula for the sum of a geometric series is $\dfrac{a}{1-r}$ where $a$ is the first term in the series and $r$ is the ratio you're multiplying by.
Here, $a = (\dfrac{1}{3})^{2\times1} = \dfrac{1}{9}$, and $r = \dfrac{1}{9}$.
So, we have $\dfrac{\dfrac{1}{9}}{1-\frac{1}{9}} = \dfrac{1}{9-1} = \dfrac{1}{8}$. 
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