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#### lfdahl

##### Well-known member

- Nov 26, 2013

- 719

$$\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\left(n\prod_{i=0}^{n}\frac{1}{j+i}\right)$$

- Thread starter lfdahl
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- Thread starter
- #1

- Nov 26, 2013

- 719

$$\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\left(n\prod_{i=0}^{n}\frac{1}{j+i}\right)$$

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- Nov 26, 2013

- 719

Hint:

The answer is: $e-1.$

The answer is: $e-1.$

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- Nov 26, 2013

- 719

Let

\[\alpha_j(n) = \frac{1}{j(j+1)(j+2)...(j+n)}\]

and let

\[\beta_j(n) = \frac{1}{j(j+1)(j+2)...(j+n-1)}\]

Consider the difference:

\[\beta_j(n)-\beta_{j+1}(n) = \frac{1}{j(j+1)(j+2)...(j+n-1)}-\frac{1}{(j+1)(j+2)(j+3)...(j+n)} \\\\ = \frac{j+n-j}{j(j+1)(j+2)...(j+n)} \\\\ = n \alpha_j(n)\]

Now

\[\sum_{j=1}^{\infty} \alpha_j(n) = \frac{1}{n}\sum_{j=1}^{\infty}\left ( \beta _j(n)-\beta _{j+1}(n) \right )\]

is a telescoping sum, and we get (the limit of $\beta$ is zero):

\[\sum_{j=1}^{\infty} \alpha_j(n) = \frac{1}{n}\left ( \beta _1(n)-\lim_{j \to \infty}\beta _j(n) \right ) =\frac{\beta _1(n)}{n}= \frac{1}{n \cdot n!}\]

Finally, we´re able to evaluate the given double sum above:

\[\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\left (n\prod_{i=0}^{n}\frac{1}{j+i} \right ) = \sum_{n=1}^{\infty}n \sum_{j=1}^{\infty} \alpha _j(n) = \sum_{n=1}^{\infty}\frac{1}{n!} = e-1.\]