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May 10th, 2015, 09:15 AM   #1
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Convergence and Series Sum

For all fixed $p\in \mathbb{N}$, prove that the series $\sum _{1}^{\infty }\frac{1}{n(n+1)...(n+p)}$ It is convergent and compute their sum.
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May 10th, 2015, 09:32 AM   #2
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The partial fraction forms a telescoping sum. I imagine you can prove this by induction.
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May 10th, 2015, 01:23 PM   #3
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As long as p > 0, $\displaystyle \frac{1}{n(n+1)...(n+p)} < \frac{1}{n^2}$. The sum over the latter converges.
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May 10th, 2015, 01:30 PM   #4
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Quote:
Originally Posted by mathman View Post
As long as p > 0, $\displaystyle \frac{1}{n(n+1)...(n+p)} < \frac{1}{n^2}$. The sum over the latter converges.


And the sum how do I calculate ?????
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May 10th, 2015, 04:51 PM   #5
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$\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+1)...(n+p)}=\frac{1} {p\cdot p!}$
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