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I was reading about telescoping series and saw this example and was interested in its result. But I was a little confused because some of the steps to the derivation were skipped. In particular, I am interested in knowing why $r$ was removed from $u_{n}$ and how it was decomposed to yield $\frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)}$. Some explanation as to how the final result was determined would also be greatly appreciated. Thank you!

$$\sum_{k=1}^nu_n=\sum_{k=1}^n \frac{1}{k(k+1)...(k+r)} = \frac{1}{r} \sum_{k=1}^n \left( \frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)} \right) = \frac{1}{r} \left( \frac{1}{r!} - \frac{n!}{(n+r)!} \right).$$

joshual
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    For the first question, put the two fractions over a common denominator. For the second, what are you left with when the sum telescopes? – saulspatz Dec 06 '19 at 20:28
  • I think it helped me to figure out what I was looking for. I see that what they did is this: $$\sum_{k=1}^nu_n=\frac{1}{r}\sum_{k=1}^n \frac{r+k-k}{k(k+1)...(k+r)}$$ and then they proceeded to decompose the fraction into the following: $$\frac{k+r}{k(k+1)...(k+r)}-\frac{k}{k(k+1)...(k+r)}$$ which simplifies to $$\frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)}$$ – joshual Dec 07 '19 at 01:25

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