# Does the following series converge or diverge?

$$\sum_n\frac{2n-1}{n!}$$

I used the ratio test here and got the lim as $n \to \infty$ to be $0$. Therefore, I assumed that the series converges. However, my textbook says that it diverges.

How do I do this problem?

• Do you mean "sequence" or "series"? That is, are you asking if the terms converge or if their sum converges? In fact, they both do. – MPW Mar 13 '14 at 2:30
• the question says "in the following series" :P – nshah Mar 13 '14 at 2:34
• I saw that, but OP did not write a series, he wrote a generic term. It has been my experience that students not infrequently misuse "series" when "sequence" is intended. I wanted to be sure what OP really intended. :) – MPW Mar 13 '14 at 2:38
• oh ok sorry haha – nshah Mar 13 '14 at 20:55

$$\sum_{n=1}^{\infty} \frac{2n-1}{n!}=\left(\sum_{n=1}^{\infty} \frac{x^{2n-1}}{n!}\right)'\left.\right|_{x=1}=\left(\frac{e^{x^2}-1}{x}\right)'\left.\right|_{x=1}=e+1$$
As John has pointed out, your textbook is wrong: the series converges. In fact, it's easy to find its value. For example, $\sum _{n=1}^\infty \frac{2n-1}{n!} = 2 \sum _{n=1} ^\infty \frac{1}{(n-1)!} - \sum _{n=1}^\infty \frac{1}{n!} = 2e - e + 1 = e + 1$.