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The series is, $$\sum \limits_{n = 2}^\infty \frac{1}{(n-1)n(n+1)} \space(a)$$

By partial fractions I've got,

$$\sum \limits_{n = 2}^\infty \frac{1}{2(n-1)}-\frac{1}{n}+\frac{1}{2(n+1)}$$

The book says that the series in $(a)$ is a Mengoli series, but I can't see how. A Mengoli series have the form of $\sum \limits_{k} u_{k}-u_{k+1}$, but I don't see any similarities. Thanks.

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@MarkBennet Made a sign error lol I see what I did wrong – Ethan Mar 22 '14 at 18:10
up vote 4 down vote accepted

Try $$\left(\frac 1{2(n-1)}-\frac 1{2n}\right)-\left(\frac 1{2n}-\frac 1{2(n+1)}\right)=$$$$\frac 1{2n(n-1)}-\frac 1{2(n+1)n}=$$$$\frac 1{(n-1)n(n+1)}$$ which verifies the partial fraction decomposition.

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Thanks! I wouldn't have seen the trick. – João Mar 22 '14 at 18:22
@João This is probably one you'll see again in some form. – Mark Bennet Mar 22 '14 at 18:32

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