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We have the following infinite sum:

$1+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+\dots$

What technique can I use to show that this is equal to $\sinh(1)$ (without to know that it is going to be $\sinh(1)$ at the start)? I only know that

$e^x = \sum\nolimits_{n=0}^\infty \frac{x^n}{n!}$ which probably don't help.

Maadhav
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Maddude
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  • You mean $\displaystyle e^x = \sum\nolimits_{n=0}^\infty \frac{x^n}{n!}$, I hope. If you want to show something about $\sinh(1)$, and don't know (and don't start at) the definition of $\sinh(x)$, you're beyond help, sorry! –  Dec 20 '17 at 16:05
  • On the opposite, it helps a lot. Consider the development of $e^{-x}$ and see how you can obtain the given sum. –  Dec 20 '17 at 16:08
  • Consider watching Herb Gross's explanation of the hyperbolic functions on youtube from his "Calculus Revisited" series from MIT opencourseware. In the video he gives intuition as to the geometric and algebraic definitions of the hyperbolic sine as well as gives proofs of identities including the series representation. – JMoravitz Dec 20 '17 at 16:14
  • Related: https://math.stackexchange.com/questions/12340/what-is-limit-of-sum-limits-n-0-infty-frac12n, https://math.stackexchange.com/questions/1708900/sum-of-sum-limits-n-0-infty-frac1kn – Hans Lundmark Dec 20 '17 at 19:31

2 Answers2

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The technique you should use is, "start from a sequence you know to understand a sequence you don't."

Okay, so starting with mystery sequence:

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

Well, it kinda looks like $e^1$, except without the even-numbered parts. How to get rid of those? Well, we can subtract them:

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

The $2$ comes from double-counting when $n$ is odd. When we divide and replace the known sequences with $e$, we're left with:

$$\frac{e^1 - e^{-1}}{2} = \sinh{1}$$

Larry B.
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I think you meant $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ but from here you can consider that $\sinh 1 = \frac{1}{2}(e^1 + e^{-1})$, and consider the series you get for $e^1$ and $e^{-1}$, and the answer should be clear from there.

B. Mehta
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  • what if I don't know that its going to be $sinh(1)$ from the start? – Maddude Dec 20 '17 at 16:05
  • Seeing that the denominators are factorials, I'd personally start by writing down the series expansions for e^x, sin, cos, sinh and cosh and trying to match something up. Further, the numerators are constant and don't change sign, indicating e^1, sinh(1) and cosh(1) are relevant. – B. Mehta Dec 20 '17 at 16:08
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    @Maddude If you don't know anything, you can't do anything (except learning). –  Dec 20 '17 at 16:09