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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
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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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  • $\begingroup$ 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
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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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    $\begingroup$ 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
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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
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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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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 $f(x) = ae^{\lambda_1x}+be^{\lambda_2x}$

$\lambda^2-1=0$

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

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

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

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

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

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

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  • $\begingroup$ 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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