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