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
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: $$\sum_{n=1}^{\infty}\frac{2}{e^n}-\sum_{n=1}^{\infty}\frac{1}{ne^n}$$
The sum on the left converges via geometric test and the sum on the right clearly converges via the ratio test. Again, your method is by no means wrong I am simply offering another approach.
