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I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$

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

I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$ so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$

But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?

Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?

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  • $\begingroup$ How did you get $\frac{1}{n!} = \frac{1}{n!} - 1$? $\endgroup$ – mattos Jan 9 '15 at 18:58
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    $\begingroup$ @Mattos, he added and subtracted the $n = 0$ term, $\frac{1}{0!}$, which is by convention $1$ $\endgroup$ – Charles Baker Jan 9 '15 at 18:59
  • $\begingroup$ @user52733 Ah, I didn't look at the bounds of his sum. I should have been more careful there. $\endgroup$ – mattos Jan 9 '15 at 19:00
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$$\begin{align} \text{Your sum} &= \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} \dots \\ \\ {\rm e} &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}\dots \end{align}$$

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Hint: if $m=n+1$, $\displaystyle \sum_{n=1}^\infty \dfrac{1}{(n+1)!} = \sum_{m=2}^\infty \dfrac{1}{m!}$.

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Use the telescope rule after your third line, that is:$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{n!}} - \frac{1}{{(n + 1)!}}} \right)} = 1$$ ;)

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$$f(x) = x e^{x} = \sum_{n=1}^{\infty} \frac{x^{n}}{(n-1)!} $$

Integrate to get

$$\int_0^x dt \, t e^t = \sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)(n-1)!} $$

Integrate by parts...

$$\int_0^x dt \, t e^t = x e^x - \int_0^x dt \, e^t = (x-1) e^x +1$$

Plug in $x=1$ on both sides to get

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

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  • $\begingroup$ You can also do this "the other way": $g(x) = (e^x - 1)/x = \sum_{n=0}^\infty \frac{x^n}{(n+1)!}$. Then $g'(1) = \sum_{n=1}^\infty \frac{n}{(n+1)!}$. We have $g'(x) = \frac{xe^x - e^x + 1}{x^2}$ so $g'(1) = \frac{e - e + 1}{1} = 1$. $\endgroup$ – aes Jan 9 '15 at 19:08
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Change the summation variable to $k = n+1$. $$\sum_1^\infty \frac{1}{(n+1)!} = \sum_2^\infty \frac{1}{(k)!} = e - \frac{1}{0!} - \frac{1}{1!} = e-2 $$

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

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$$ \sum\limits_{n = 1}^{ + \infty } {\frac{1}{{\left( {n - 1} \right)!\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}} = \sum\limits_{n = 0}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}} = 1 $$

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  • $\begingroup$ Proof of the last equality would be useful. $\endgroup$ – apnorton Jan 12 '15 at 19:57
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Consider $$f(x)=\sum^{\infty}_{n=1}\frac{x^{n+1}}{(n-1)!(n+1)}\Rightarrow f'(x)=\sum^{\infty}_{n=1}\frac{x^{n}}{(n-1)!}=x\sum^{\infty}_{n=1}\frac{x^{n-1}}{(n-1)!}=x\sum^{\infty}_{n=0}\frac{x^{n}}{n!}=xe^x$$ Therefore $$f(x)=\int xe^x\,dx=xe^x-e^x+c$$ where $c=f(0)+e^0-0\cdot e^0=1$ so at $x=1$ you get your sum $$f(1)=\sum^{\infty}_{n=1}\frac{1}{(n-1)!(n+1)}=1\cdot e^1-e^1+c=c=1$$

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To evaluate the second series note: $$ e=\sum_{n=0}^\infty\frac{1}{n!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$$ $$ \sum_{n=1}^\infty\frac{1}{(n+1)!}=\frac{1}{2!}+\frac{1}{3!}+\dots$$

Thus the second series is just $e-2$. If you combine the two series you get $1$.

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