How would I compute:
$$\sum_{n=2}^\infty \frac{1}{n^2 - n} \cdot n$$
Hints or step by step process would be the most helpful.
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2$\begingroup$ Is the expression $(1 / (n^2 - n)) \cdot n$ correct? It looks a little off. $\endgroup$– SamMar 16, 2013 at 18:09
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1$\begingroup$ This is same as the harmonic series which diverges... You sure the question is correct? $\endgroup$– Gautam ShenoyMar 16, 2013 at 18:11
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$\begingroup$ I believe so. What exactly seems to be off about it? It's supposed to be an expected value. $\endgroup$– JohnMar 16, 2013 at 18:15
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$\begingroup$ @John Multiplying by $n$ at the end like that seemed strange, and I was afraid that I had edited the expression too hastily. Given that the sum turned out to be the harmonic series, it doesn't seem to be off at all. $\endgroup$– SamMar 16, 2013 at 18:31
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1 Answer
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$$\sum_{n\geq 2} n/(n^2-n)=\sum_{n\geq 2} n/(n(n-1))=\sum_{n\geq 2} 1/(n-1)=1+1/2+1/3+...=\infty$$ is divergent harmonic series
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1$\begingroup$ @John You could make the substitution $m = n - 1$. $\endgroup$– SamMar 16, 2013 at 18:33
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$\begingroup$ Why is the final answer infinity? Eventually $\frac1{large number}$ becomes very close to 0. My calculator (casio fx-570 es plus) calculated $S_{999} = 7.48$ and then $S_{9999} = 9.79$. $\endgroup$– OzzyMar 16, 2013 at 20:11
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$\begingroup$ see en.wikipedia.org/wiki/Harmonic_series_(mathematics) $\endgroup$– Adi DaniMar 16, 2013 at 20:16