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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    $\begingroup$ What tests do you know for convergence? $\endgroup$ – Qiaochu Yuan Oct 20 '10 at 21:32
  • $\begingroup$ 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... $\endgroup$ – 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.


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.

  • $\begingroup$ 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? $\endgroup$ – Robert Smith Oct 21 '10 at 0:18
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    $\begingroup$ $\sinh(x)=\frac{\exp(x)-\exp(-x)}{2}$. Also, [this](http://functions.wolfram.com/ElementaryFunctions/Sinh/) gives this. $\endgroup$ – J. M. is a poor mathematician Oct 21 '10 at 1:21
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    $\begingroup$ @Robert: It is the Taylor series for sinh evaluated at 1. $\endgroup$ – GEdgar Jun 16 '12 at 16:28
  • $\begingroup$ Thanks. I didn't know that. $\endgroup$ – 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.

  • 6
    $\begingroup$ I don't see the point of doing this when proving that the series for e converges is exactly as hard. $\endgroup$ – Qiaochu Yuan Oct 20 '10 at 22:23
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    $\begingroup$ In short, $e=\cosh\;1+\;sinh\;1$. :) $\endgroup$ – J. M. is a poor mathematician Oct 21 '10 at 0:19
  • $\begingroup$ @ Qiaochu Yuan. You make a valid point. I guess the Ratio test(which Robert Smith has mentioned) is the rigorous answer :). $\endgroup$ – 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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    $\begingroup$ The question was «does the series converge?» Your answer begins by asserting convergence of a series of equivalent difficulty :) $\endgroup$ – Mariano Suárez-Álvarez 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.


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

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



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


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.



This is how sometimes we can extract the exact closed form of some series :

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

The radius of convergence is $\infty$

$f \in C^{\infty}(\mathbb{R})$

$f''(x) = f(x)$

solving our equation, we get $f(x) = ae^{\lambda_1x}+be^{\lambda_2x}$


$\lambda \in \{-1,1\}$

$f(x) = ae^{-x}+be^{x}$

$f(0)= 0 \implies a+b=0 \implies a=-b \implies f(x) = a(e^{x}- e^{-x})$

$f'(0) = 1 \implies a=\frac{1}{2}$

$f(x) = \frac{e^{x}-e^{-x}}{2}$

so our series $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$ does converge to $\sinh(1)$

  • $\begingroup$ This only proes that if the series converges, then its sum is $\sinh(1)$. $\endgroup$ – José Carlos Santos Jul 26 '18 at 20:24
  • $\begingroup$ Yes that is true, but I wanted to say, regarding to other answers, that I would find the limit. So I didn't repeat any other done process here, that is namely true. Anyway that is the same, effectively the sum itself implies the convergence of that series to $\sinh(x)$. So you had not always to prove the convergence of series, since it does have a closed form. the closed form itself implies the convergence, that is so understandable. $\endgroup$ –  Ахмед Jul 26 '18 at 20:26
  • $\begingroup$ your implication is not true for closed form and finite values. that does mean, the sum does converge for all $x\in \mathbb{R}$ to $\sinh(x)$ which is finite. $\endgroup$ –  Ахмед Jul 26 '18 at 20:33

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