Ways of showing \sum_{n=1}^\infty\ln(1+\frac{1}{n}) to be divergent

amolent3u 2022-01-03 Answered
Ways of showing \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\ln{{\left({1}+{\frac{{{1}}}{{{n}}}}\right)}}}\) to be divergent

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Expert Answer

Philip Williams
Answered 2022-01-04 Author has 3715 answers
Notice the following:
\(\displaystyle{\log{{\left({1}+{\frac{{{1}}}{{{n}}}}\right)}}}={\log{{\left({\frac{{{n}+{1}}}{{{n}}}}\right)}}}={\log{{\left({n}+{1}\right)}}}-{\log{{\left({n}\right)}}}\)
Hence
\(\displaystyle{\sum_{{{k}={1}}}^{{n}}}{\log{{\left({1}+{\frac{{{1}}}{{{k}}}}\right)}}}={\log{{\left({n}+{1}\right)}}}\rightarrow\infty\)
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Corgnatiui
Answered 2022-01-05 Author has 792 answers
This is a special case of: Suppose \(\displaystyle{f{{\left({1}\right)}}}={0},{f}'{\left({1}\right)}{>}{0}\). Then \(\displaystyle\sum{f{{\left({1}+{\frac{{{1}}}{{{n}}}}\right)}}}=\infty\)
Proof: From the definition of the derivative (no Taylor necessary), we have
\(\displaystyle{\frac{{{f{{\left({1}+{h}\right)}}}-{f{{\left({1}\right)}}}}}{{{h}}}}={\frac{{{f{{\left({1}+{h}\right)}}}}}{{{h}}}}{>}{\frac{{{f}'{\left({1}\right)}}}{{{2}}}}\)
for small \(\displaystyle{h}{>}{0}\). Thus
\(\displaystyle{f{{\left({1}+{\frac{{{1}}}{{{n}}}}\right)}}}{>}{\frac{{{f}'{\left({1}\right)}}}{{{2}}}}\cdot{\frac{{{1}}}{{{n}}}}\)
for large \(\displaystyle{n}\). By the comparison test we're done.
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Vasquez
Answered 2022-01-11 Author has 8850 answers

\(\sum_{n=1}^m\log(\frac{n+1}{n})=\sum^m_{n=1}(\log(n+1)-\log n)=\log(m+1)\)
The partial sums clearly diverge. Alternatively using the cauchy condensation test the series converges iff the series
\(\sum^\infty_{n=1}2^n\ln(1+\frac{1}{2^n})\)
converges. The transformed series diverges since the terms don't go to zero and so the original series diverges.

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1. \(\displaystyle{\log{{\left({A}{B}\right)}}}={\log{{\left({A}\right)}}}+{\log{{\left({B}\right)}}}\)
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6. \(\displaystyle\sqrt{{{\ln{{\left({A}\right)}}}}}={\ln{{\left({A}^{{{\left(\frac{{1}}{{2}}\right)}}}\right)}}}\)
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