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3 votes
Why does this series converge?

∑ (n+1)/[(n)(n+2)(n+3)]
n=1

User MMiroslav
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1 Answer

4 votes

Compare the series to the convergent series,


\displaystyle\sum_(n=1)^\infty\frac1{n^2}

By the limit comparison test, the given series converges because


\displaystyle\lim_(n\to\infty)\frac{(n+1)/(n(n+2)(n+3))}{\frac1{n^2}}=\lim_(n\to\infty)(n(n+1))/((n+2)(n+3))=1

User Biomehanika
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