Calc6.1

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Discuss the convergence of \sum _{{n=1}}^{{\infty }}{\frac  {1}{n+1}}.

Since the terms of this sequence are positive and decreasing, we can use the integral test to see if the series converges. If the integral

\int _{{1}}^{{\infty }}{\frac  {1}{x+1}}\,dx\,

converges or diverges, then this sum will do the same.

\int _{{1}}^{{\infty }}{\frac  {1}{x+1}}\,dx=\ln |n+1|{\bigg |}_{{1}}^{{\infty }}=\infty \,

Thus, this integral diverges. By the integral test, this series also diverges.


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