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

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?

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    $\begingroup$ 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. $\endgroup$ – MPW Mar 13 '14 at 2:30
  • $\begingroup$ the question says "in the following series" :P $\endgroup$ – nshah Mar 13 '14 at 2:34
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    $\begingroup$ 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. :) $\endgroup$ – MPW Mar 13 '14 at 2:38
  • $\begingroup$ oh ok sorry haha $\endgroup$ – nshah Mar 13 '14 at 20:55
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Your textbook is wrong. This series converges. Your reasoning is correct. (As a textbook author, I sympathize with the person who made the mistake; as a teacher, I sympathize with you as a student, and applaud you for thinking your reasoning was good!)

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  • $\begingroup$ Alright thanks. I assumed i was right because i attempted the problem many times and kept getting the same answer. $\endgroup$ – nshah Mar 13 '14 at 2:23
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You can also see that

$$\sum_{n=1}^{\infty} \frac{2n-1}{n!}=\left(\sum_{n=1}^{\infty} \frac{x^{2n-1}}{n!}\right)'\left.\right|_{x=1}=\left(\frac{e^{x^2}-1}{x}\right)'\left.\right|_{x=1}=e+1$$

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

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Any polynomial function grows slower than a factorial function, so the limit must converge.

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