Calc6.49

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Discuss the convergence or divergence of the series \sum _{{k=2}}^{{\infty }}\ln \left({\frac  {k(k+2)}{(k+1)^{2}}}\right).

This is another series that does not, at first, look like a telescopic series. But let us separate the log into four logs using the laws of logarithms.

S=\sum _{{k=2}}^{{\infty }}\ln \left({\frac  {k(k+2)}{(k+1)^{2}}}\right)=\sum _{{k=2}}^{{\infty }}[\ln k-\ln(k+1)-[\ln(k+1)-\ln(k+2)]]=\ln 2-\lim _{{k\rightarrow \infty }}\ln(k+1)-\ln 3+\lim _{{k\rightarrow \infty }}\ln(k+2)\,

This has a value of +\infty and another value of -\infty so it is unclear what the sum is. Let us take these four logs and put them back together as one.

S=\lim _{{k\rightarrow \infty }}\ln {\frac  {2(k+2)}{3(k+1)}}=\ln {\frac  {2}{3}}\,


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