# Comparison Test for $\sum_{n=1}^{\infty}\frac{2n-1}{ne^n}$

I have this math problem where I have to show that a sum converges. Is this correct? $$\sum_{n=1}^{\infty}\frac{2n-1}{ne^n}$$ I chose $$\sum_{n=1}^{\infty}\frac{2n}{ne^n}$$ to compare it to.

This simplifies to $$\sum_{n=1}^{\infty}\frac{2}{e^n}$$, which is the same as $$\sum_{n=1}^{\infty}2(\frac{1}{e})^n$$.

Since $$\frac{1}{e}$$ is $$< 1$$; $$\sum_{n=1}^{\infty}\frac{2n-1}{ne^n}$$ converges

• Yes, it is correct. Commented Nov 17, 2014 at 4:07
• Correct and perhaps full credit for your context (10 years ago... not sure how it got bumped today). I appreciate an extra indication around the series being always positive, tho: comparison showing it's less than a convergent series doesn't necessarily work if you have a sequence of negative numbers. Commented Sep 4 at 22:38

Your method is correct, however, using the comparison test is not necessary. You can expand the original fraction into $$\frac{2}{e^n}-\frac{1}{ne^n}$$ and apply the sum rule leaving us with: $$\sum_{n=1}^{\infty}\frac{2}{e^n}-\sum_{n=1}^{\infty}\frac{1}{ne^n}$$