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I was expanding on specific telescopic series $$\sum_{n=1}^\infty \frac{1}{n(n+1)}= 1$$ $$\sum_{n=1}^\infty \frac{1}{n(n+2)}= \frac{3}{4}$$ $$\sum_{n=1}^\infty \frac{1}{n(n+3)}= \frac{11}{18}$$ It seems that as $c$ increases to $\infty$, the series converge to a certain value. Is there a method to find whether these series converge and what the value is, where $c\geq 1$? $$\sum_{n=1}^\infty \frac{1}{n(n+c)}=$$

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    $\begingroup$ It certainly converges for all $c$ other than negative integers. But the telescoping only works for $n$ a positive integer. $\endgroup$ Jan 20 at 2:08
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    $\begingroup$ $\sum \frac{1}{n^2}$ converges so $\sum\frac{1}{n(n+c)}$ will certainly converge if $c \geq 1$ $\endgroup$
    – layabout
    Jan 20 at 2:10
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    $\begingroup$ Use partial fractions to expand ${1 \over n(n+c)}$ (assuming $c$ is a positive integer). I think the answer is ${1 \over c} ({1 \over 1} + {1 \over 2}+\cdots + {1 \over c})$. $\endgroup$
    – copper.hat
    Jan 20 at 2:10
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    $\begingroup$ When $c>0$ is an integer, the series converges to $\frac1cH_c,$ when $$H_n=1+\frac12+\cdots+\frac1n$$ is the harmonic series. Since $H_n\sim\ln n,$ your limit is $0.$ This is pretty easy to see directly without the closed formula, I’d guess, allowing real $c$ in the limit. $\endgroup$ Jan 20 at 2:13

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For $c \in \Bbb Z_+$, you can continue your telescoping argument to show that $$ \sum_{n = 1}^\infty \frac{1}{n(n + c)} = \frac{H_c}{c} $$ where $$ H_c = \sum_{j = 1}^c \frac{1}{j} $$ is the $c$-th harmonic number. Since $H_c \sim \gamma + \ln(c)$, $H_c / c \to 0$ as $c \to \infty$.

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  • $\begingroup$ Could you explicitly show how to expand the telescoping argument? $\endgroup$ Feb 4 at 23:18

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