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!

Your method is fine indeed
$$\sum_{k=5}^\infty \frac1{3^k}=\sum_{k=0}^\infty \frac1{3^k}-\sum_{k=0}^4 \frac1{3^k}=\frac32 - \sum_{k=0}^4 \frac1{3^k}=\frac32-1-\frac13-\frac19-\frac1{27}-\frac1{81}=\frac1{162}$$
As an alternative, following the clever method suggested by lulu in the comment, we have
$$\sum_{k=5}^\infty \frac1{3^k}=\sum_{j=0}^\infty \frac1{3^{j+5}}=\frac1{3^5}\sum_{j=0}^\infty \frac1{3^{j}}=\frac1{3^5}\cdot\frac32=\frac1{2\cdot 3^4}=\frac1{162}$$
