- #1

Euler2718

- 90

- 3

## Homework Statement

Show that the sequence of partial sums

[tex] s_{n} = 1+\sum_{i=1}^{n} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right)\right) [/tex]

converges, with [itex]n\in \mathbb{N}\cup \{0\}[/itex]

## Homework Equations

## The Attempt at a Solution

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So we want to find

[tex] \lim_{n\to\infty} s_{n} = \lim_{n\to\infty} \left(1+ \sum_{i=1}^{n} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right)\right) \right) = 1 + \sum_{i=1}^{\infty} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) \right) [/tex]

I haven't had much experience in a summation of a product, so my best guess was to determine convergence of the product first. Rewriting [itex]\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) = \prod_{k=1}^{\infty}\left( 1 + \frac{2-k}{2k}\right)[/itex] the problem then becomes [itex]\sum_{k=1}^{\infty}\frac{2-k}{2k}[/itex] which diverges by the limit comparison test. If I recall correctly this implies the product should diverge to 0? So the infinite sum of this is just 0, and the limit of partial sums is then 1? However wolfram alpha begs to differ and it gives me the answer of 8. How should I have proceeded?