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Let $ m \in \mathbb{N} $. Find to what does the series $ \sum\limits_{n=1}^{\infty}\frac{1}{n(n+m)} $ converge to

My attempt: ( I did as discussed in Infinite Series $\sum 1/(n(n+1))$ )

Let $ N>m $.
$ \begin{align} S_N & = \sum\limits_{n=1}^{N}\frac{1}{n(n+m)} = \frac{1}{m}\sum_{n=1}^N \left(\dfrac1n - \dfrac1{n+m}\right) = \frac{1}{m}( \sum_{n=1}^N \dfrac1 n - \sum_{n=1}^N \dfrac1{n+m}) = \frac{1}{m}(\sum_{n=1}^N \dfrac1n - \sum_{n=m+1}^{N+m} \dfrac1{n} )\\ & = \frac{1}{m}(\sum_{n=1}^m \dfrac1n + \sum_{n=m+1}^N \dfrac1n - \sum_{n=m+1}^N\dfrac1n - \sum_{n=N+1}^{N+m}\dfrac1n ) = \frac{1}{m}(\sum_{n=1}^m \dfrac1n - \sum_{n=N+1}^{N+m}\dfrac1n ) ~~\text{ [ from here I got stuck] } \end{align} $

hazelnut_116
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    As $N\to\infty$ the second sum is not relevant anymore in the infinite sum. The first sum is the harmonic series, there is no specific term to express that. So if there is no more $N$ inside your formula you have the convergence value $\frac{1}{m}\sum_{n=1}^m \frac{1}{n}$ and you are finished. – LegNaiB May 10 '21 at 20:20
  • @LegNaiB You are right, I missed this somehow. – hazelnut_116 May 10 '21 at 20:23
  • You have to differentiate between the $n$ inside the finite sum which is always $\leq m$ and the $N$ going to infinity – LegNaiB May 10 '21 at 20:24

1 Answers1

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You are almost done. We have $$\begin{align} S_N &= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = N+1}^{N+m} \frac{1}{n} \\ &= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N} \\ &= \frac{1}{m} H_m - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N}, \end{align}$$

where I have used $H_m = \sum_{n=1}^m \frac{1}{n}$ to represent the $m^{\rm th}$ harmonic number. hence for a fixed $m$, as $N \to \infty$,

$$\lim_{N \to \infty} S_N = \frac{H_m}{m} - \frac{1}{m} \lim_{N \to \infty} \sum_{n=1}^m \frac{1}{n + N}.$$ But since the remaining sum contains only finitely many terms, we can interchange the order of summation and limit to obtain

$$\lim_{N \to \infty} S_N = \frac{H_m}{m} - \frac{1}{m} \sum_{n=1}^m \lim_{N \to \infty} \frac{1}{n + N} = \frac{H_m}{m} - \frac{1}{m} \sum_{n=1}^m 0 = \frac{H_m}{m}.$$

heropup
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