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Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$

$$ S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\ =\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{1}{6.3!}+\dots\bigg]=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big) $$ How do I proceed further as I am stuck with the last infinite series.

Kenta S
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Sooraj S
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4 Answers4

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Hint: Take two derivation of both side of $$\dfrac{e^x-1-x-\dfrac12x^2}{x}=\sum_{n=2}^{\infty}\dfrac{x^n}{(n+1)!}$$ then let $x=1$.

Nosrati
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$$\dfrac{\binom n2}{(n+1)!}=\dfrac{n(n-1)}{2(n+1)!}$$

As $n(n-1)=(n+1)n-2(n+1)+2$

$$ \dfrac{\binom n2}{(n+1)!}=\dfrac12\cdot\dfrac1{(n-1)!}-\left(\underbrace{\dfrac1{n!}-\dfrac1{(n+1)!}}\right)$$

The terms under brace telescopes

Use $e^y=\displaystyle\sum_{r=0}^\infty\dfrac{y^r}{r!}$ for the first term

See also : Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$

lab bhattacharjee
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$$S=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}=\sum_{n=0}^\infty\frac{1}{2(n+3)(n)!}$$ $$=\frac{1}{2}\sum_{n=0}^\infty\int_{0}^{1}\frac{x^{n+2}}{n!}dx=\frac{1}{2}\int_{0}^{1}{x^2e^{x}}dx=\frac{e}{2}-1$$

E.H.E
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First, it must be: $$S=\sum_{n=2}^\infty\frac{^nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}=\\ \color{blue}{=\sum_{n=2}^\infty\frac{(n+1)-n}{2(n+1)(n-2)!}=\frac12\left[\sum_{n=2}^\infty\frac{1}{(n-2)!}-\sum_{n=2}^\infty\frac{n}{(n+1)(n-2)!}\right]=}\\ =\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{\color{red}3}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{\color{red}3}{4.1!}+\dots\Big).$$ Second, evaluating the second series is more difficult: $$\sum_{n=2}^\infty\frac{n}{(n+1)(n-2)!}=\sum_{n=2}^\infty\frac{n^2(n-1)}{(n+1)!}=\\ \sum_{n=2}^\infty\frac{(n+1)(n^2-2n+2)-2}{(n+1)!}=\sum_{n=2}^\infty\frac{(n+1)n(n-1)-(n+1)n+2(n+1)-2}{(n+1)!}=\\ \sum_{n=2}^\infty\left[\frac{1}{(n-2)!}-\frac{1}{(n-1)!}\right]+2\sum_{n=2}^\infty\left[\frac{1}{n!}-\frac{1}{(n+1)!}\right]=\\ \left[\frac1{0!}-\frac1{1!}+\frac1{1!}-\frac1{2!}+\frac1{2!}-\frac1{3!}+\cdots\right]+2\left[\frac1{2!}-\frac1{3!}+\frac1{3!}-\frac1{4!}+\frac1{4!}-\frac1{5!}+\cdots\right]=2.$$ Thus, following lab bhattacharjee's method is more efficient.

farruhota
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