Find all values of $x$ for which the series converges, and find the sum of the series for those values of $x$. $$e^{-11x}+e^{-22x}+e^{-33x}+e^{-44x}+e^{-55x}+\cdots$$
I figured that I can rewrite this as $$\sum_{n=1}^{\infty}(e^{-11x})^{n}$$
I figured that $r = e^{-11x}$ and $a=1$
I had to solve for $\left | e^{-11x}\right |<1$, so $x>0$
Would the sum be?: $$\frac{e^{-11x}}{1-e^{-11x}}$$
Thanks
As $\exp$ is positive (unless you have a complex variable), you can drop the $|.|$. Then the condition is $$ e^{-11x} < 1 $$ As $\log$ increases, this is equivalent to $$ -11x < 0 \iff x>0 $$under these conditions, the sum value is what you write.