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I have a geometric series that looks like $$ f(m) = \sum_{r=1}^{m}\frac{r+1}{n-r-1} \\ = \frac{2}{n-2} + \frac{3}{n-3} + \frac{4}{n-4} + \cdots + \frac{m+1}{n-(m-1)} \\ $$ I wrote out a few terms to see if I can identify a simplification but I couldn't. What kind of geometries series is this, and how can I simplify to get a closed form expression

roulette01
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1 Answers1

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I'm assuming $n$ is fixed. Note that in order for the sum to be well-defined, we must have $n \leq m$.

A closed form is unlikely, but: \begin{align*} f(m) &= \sum_{r=1}^m \frac{r+1}{n - (r + 1)} \\ &= \sum_{r=1}^m \left(\frac{n}{n - (r + 1) }- 1\right) \\ &= n\sum_{r=1}^m \frac{1}{n - (r + 1)} - m \\ &= n(H_{n-2} - H_{n-m-2}) - m \end{align*} where $H_k$ is the $k^\text{th}$ harmonic number.

Clement Yung
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  • Is $H_{n-2} - H_m$ right? I got the following: $$ \bigg[n \bigg( \frac{1}{n-2} + \frac{1}{n-3} + \cdots + 1] + [\frac{1}{-1} + \frac{1}{-2} \cdots \frac{1}{n-(m+1)}\bigg)\bigg] -m \ = \bigg[n \bigg( \frac{1}{n-2} + \frac{1}{n-3} + \cdots + 1 \bigg) - \bigg(\frac{1}{1} + \frac{1}{2} \cdots \frac{1}{|n-(m+1)|}\bigg)\bigg] -m $$

    Oh wait, are you assuming $m < n$?

     – roulette01 May 09 '20 at 13:17
  • @dd22205 yes I'm assuming $m \leq n$. I'll add that in my answer. If $m \geq m + 1$, then your sum can't be well-defined as you'll have a $\frac{1}{0}$ term. – Clement Yung May 09 '20 at 13:23
  • Ah yes, that's right. Shouldn't it be $H_{n-2} - H_{n-(m)}$? – roulette01 May 09 '20 at 13:25
  • @dd22205 you're right, I've edited my answer. – Clement Yung May 09 '20 at 14:49
  • My bad. There's it's actually $H_{n-m-2}$. Went in the wrong direction. – roulette01 May 09 '20 at 16:02