Calc6.47

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Discuss the convergence or divergence of the series \sum _{{k=1}}^{{\infty }}\left({\frac  {1}{k+7}}-{\frac  {1}{k+8}}\right).

This is a telescopic series. Every term of this series cancels out part of the previous term so finding the sum is very easy.

\sum _{{k=1}}^{{\infty }}\left({\frac  {1}{k+7}}-{\frac  {1}{k+8}}\right)={\frac  {1}{8}}-{\frac  {1}{9}}+{\frac  {1}{9}}-{\frac  {1}{10}}+{\frac  {1}{10}}-{\frac  {1}{11}}+...={\frac  {1}{1+7}}-\lim _{{k\rightarrow \infty }}{\frac  {1}{k+7}}={\frac  {1}{8}}\,


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