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Determine whether this series converges: \sum _{{k=1}}^{\infty }{\frac  {1}{k(k+1)}}\,

Use partial fraction decompostition to write the first n terms of this sum as:

\sum _{{k=1}}^{n}\left({\frac  {1}{k}}-{\frac  {1}{k+1}}\right)\,

=(1-{\frac  {1}{2}})+({\frac  {1}{2}}-{\frac  {1}{3}})+({\frac  {1}{3}}-{\frac  {1}{4}})+...+({\frac  {1}{n-1}}-{\frac  {1}{n}})+({\frac  {1}{n}}-{\frac  {1}{n+1}})\,

Notice that all of the middle terms cancel out.

=1-{\frac  {1}{n+1}}\,

Take the limit as n goes to infinity of this sum.

\sum _{{k=1}}^{\infty }{\frac  {1}{k(k+1)}}=\lim _{{n\rightarrow \infty }}\sum _{{k=1}}^{n}{\frac  {1}{k(k+1)}}=\lim _{{n\rightarrow \infty }}\left(1-{\frac  {1}{1+n}}\right)=1\,

So the series converges to 1.


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