<s>[INST] Finding sum of a geometric series I am asked to find the summation of $1/3^n$ from $n=5$ to infinity.
I have done the calculation: $1/(1-r)$, for $r=1/3$, and received $1.5$. As this summation starts from $5$, I subtracted $3^0, 3^-1, 3^-2, 3^-3$ and $3^-4$ from $1.5$ and got $6.17e-3$. However, apparently this answer is wrong, and so is the answer 0. 
I appreciate any help, 
thank you!
 [/INST] $$1,\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\frac{1}{3^4} \ldots$$
You need to calculate $$S_{\infty}-S_5$$
$$=\frac{1}{1-\frac{1}{3}}-\frac{1(1-\frac{1}{3^5})}{1-\frac{1}{3}}$$
$$=\frac{\frac{1}{3^5}}{\frac{2}{3}}$$
$$=\frac{1}{2\cdot{3^4}}$$
$$=\frac{1}{162}$$
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