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

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

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Philip Williams
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
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.
Vasquez

$$\sum_{n=1}^m\log(\frac{n+1}{n})=\sum^m_{n=1}(\log(n+1)-\log n)=\log(m+1)$$
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.