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

This question appeared while solving other alternate sums: $\sum\limits_{n=1}^\infty a_{n+1}-a_n$ and the thing is that it converges searching via: $\lim\limits_{n\to\infty} S_n=\lim\limits_{n\to\infty}a_{n+1}-a_1$

The problem is to express the first sum as an alternate sum, so I need to find a formula for $a_n$ and the rest is easy. I've solved the others problems using partial fraction decomposition for the first expression.

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    $\begingroup$ Note that $\prod_{i=1}^n(1+i/2)=(3/2)(4/2)(5/2)...((n+2)/2)=\frac{(n+2)!}{2(2^n)}=\frac{(n+2)!}{2^{n+1}}$ $\endgroup$ – Mark May 6 '16 at 1:36
  • $\begingroup$ After using @Mark's hint, you can treat the sum using partial fraction decomposition and actually calculate its value. $\endgroup$ – Clayton May 6 '16 at 1:52
  • $\begingroup$ @Mark Yes, after Mark's hint it's really easy. Thanks! $\endgroup$ – Erick May 6 '16 at 14:51
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As @Mark pointed out, you're problem can be reexpressed as.

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

$$\sum\limits_{n=0}^{\infty}\frac{2 n!}{(n+2)!}$$

$$\sum\limits_{n=0}^{\infty}\frac{2}{(n+1)(n+2)}$$

$$2 \times \left(\sum\limits_{n=0}^{\infty}\frac{1}{n+1} - \sum\limits_{n=0}^{\infty}\frac{1}{n+2}\right)$$

Making a change of variables

$$2 \times \left(\sum\limits_{k=1}^{\infty}\frac{1}{k} - \sum\limits_{k=2}^{\infty}\frac{1}{k}\right)$$

$$2 \times \left(\sum\limits_{k=1}^{\infty}\frac{1}{k} - \left( \sum\limits_{k=1}^{\infty}\frac{1}{k} - 1 \right) \right)$$

$$2$$

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    $\begingroup$ Your last 3 expressions are problematic because each of those series diverge. Conisder partial sums instead. $\endgroup$ – zhw. May 6 '16 at 1:56
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We have $$ S=\sum_{n=0}^{+\infty} \frac{n!}{2^n \prod_{i=1}^n \left(1+\frac{i}{2} \right)}, $$ but we know $$ \prod_{i=1}^n \left(1+\frac{i}{2} \right)=\frac{\Gamma(n+3)}{2^{(n+1)}} = \frac{(n+2)!}{2^{(n+1)}}. $$ So, we have $$S=\sum_{n=0}^{+\infty} \frac{n!}{2^n \frac{(n+2)!}{2^{(n+1)}}} = 2 \sum_{n=0}^{+\infty} \frac{n!}{(n+2)!} = 2\sum_{n=0}^{+\infty} \left( \frac{1}{n+1}-\frac{1}{n+2} \right) = 2$$

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  • $\begingroup$ I think you've missed a 2, bit yeah thanks! $\endgroup$ – Erick May 6 '16 at 14:50
  • $\begingroup$ @Erick, I'm sure I did not. I'm completly sure about the result. Besides, it is cordial to upvote the right answer as a thank you. $\endgroup$ – Guilherme Thompson May 6 '16 at 15:00

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