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!

You also can use this formula, valide if $\;|x|<1$:
$$\sum_{n=N}^\infty x^n=\frac{x^N}{1-x},$$
which is a straightforward corollary of the formula for the sum of the geometric series:
$$\sum_{n=N}^\infty x^n=x^N\sum_{n=0}^\infty x^n=x^N\frac1{1-x}.$$
