# $\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$ converges?

Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$

Thought about using the limit theorem or by comparison but am so stuck. any pointers would be appreciated guys

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What tests do you know for convergence? – Qiaochu Yuan Oct 20 '10 at 21:32
Look at the absolute value of the ratio of a term in the series to the subsequent term. You should already know what the possible results imply... – Brandon Carter Oct 20 '10 at 23:35

Another way is

If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$

We have that

$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$

$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} < 2$

Thus $S_n < 2$

thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.

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I think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is obviously a number, so he doesn't have to prove that the series for e converges.

Maybe you can prove the same by using the ratio test $\lim_{n \rightarrow \infty} \displaystyle |\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\displaystyle \frac{1}{(2n+1)!}$ and $a_{n+1}=\displaystyle \frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\lim_{n \rightarrow \infty} \displaystyle \frac{1}{(2n+3)(2n+2)}$ which is 0. According to the ratio test:

If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.

Therefore, the series converges.

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Just out of the curiosity, I plugged the series into Wolfram|Alpha. It gives the very nice result of sinh(1). If anyone knows how to prove that, it would be wonderful to know it. By the way, how can I add an hyperlink in the comment section? – Robert Smith Oct 21 '10 at 0:18
$\sinh(x)=\frac{\exp(x)-\exp(-x)}{2}$. Also, [this](http://functions.wolfram.com/ElementaryFunctions/Sinh/) gives this. – J. M. Oct 21 '10 at 1:21
@Robert: It is the Taylor series for sinh evaluated at 1. – GEdgar Jun 16 '12 at 16:28
Thanks. I didn't know that. – Robert Smith Jun 16 '12 at 18:06

The series you have is

$1 + \frac{1}{3!} + \frac{1}{5!} \ldots$

If you add the even factorial terms, you get an upper bound i.e.,

$1 + \frac{1}{3!} + \frac{1}{5!} \ldots < \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}+ \frac{1}{5!} \ldots$

This can be written more compactly as

$\sum_{n=0}^\infty \frac{1}{(2n+1)!} < \sum_{n=0}^\infty \frac{1}{n!} = e^1$

Therefore the series converges.

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I don't see the point of doing this when proving that the series for e converges is exactly as hard. – Qiaochu Yuan Oct 20 '10 at 22:23
In short, $e=\cosh\;1+\;sinh\;1$. :) – J. M. Oct 21 '10 at 0:19
@ Qiaochu Yuan. You make a valid point. I guess the Ratio test(which Robert Smith has mentioned) is the rigorous answer :). – svenkatr Oct 21 '10 at 3:53

We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots$$

Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$

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The question was «does the series converge?» Your answer begins by asserting convergence of a series of equivalent difficulty :) – Mariano Suárez-Alvarez Oct 21 '10 at 14:12

Can you bound the series from above by one that you know converges? The factorials grow very fast, so you should be able to.

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To elaborate on the first answer given to this question by Ross Millikan.

$$\sum_{n=0}^\infty \frac{1}{(2n+1)!} = 1 + \sum_{n=1}^\infty \frac{1}{(2n+1)!}$$

$$< 1 + \sum_{n=1}^\infty \frac{1}{4^n} = \frac{4}{3}, \quad \textrm{ as } \frac{1}{(2n+1)!} < \frac{1}{4^n} \textrm{ for } n \ge1.$$

Hence by the comparison test the series converges.

Comparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that

$$R_N < \frac{1}{(2n+3)!} \left( 1 + \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$

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METHOD I

We may simply resort to the Basel problem and get the inequality: $$0<\sum_{k=0}^{\infty}\frac{1}{(1+2k)!}\leq\sum_{k=0}^{\infty}\frac{1}{(1+k)^2}=\frac{\pi^2}{6}$$

METHOD II

According to Taylor's expansion we have that:

$$\sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!}$$

For $x=1$ we get that the value of the series is $\sinh(1)$. The series converges.

Q.E.D.

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