# Does the series $\sum_{n=1}^{\infty}\log n - (\log n)^{n/(n+1)}$ converge?

Does the series

$$\sum_{n=1}^{\infty}\log n - (\log n)^{n/(n+1)}$$

converge?

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## 3 Answers

By AM-GM:

$$\sqrt[n+1]{ \log(n)^n} \leq \frac{n\log(n)+1}{n+1}$$

Thus

$$\log n- (\log n)^{\frac{n}{n+1}} \geq \log(n)-\frac{n\log(n)+1}{n+1}=\frac{\log(n)-1}{n+1}$$

Now, Limit Comparison test tells you that since $\sum \frac{\log n}{n}$ is divergent, this series is also divergent.

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Let $n \geq 3$ then

$$\log(n) - \log(n)^{n/(n+1)} = \frac{\log(n)^{n/(n+1)}}{n+1} (n+1) \left(\log(n)^{1/(n+1)} - 1 \right) \geq \frac{\log(\log(n))}{n+1}$$

Since $\log(\log(n)) \to \infty$ and $\sum 1/(n+1)$ diverges, this series itself diverges.

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Warning, ugly calculation ahead.

We compare with the harmonic series. We will use L'Hospital's Rule, so change the variable to $x$. We look at the behaviour of $$\frac{1-(\log x)^{-1/(x+1)}}{\frac{1}{x\log x}}$$ for large $x$.

The derivative of the bottom is $-\dfrac{1+\log x}{x^2\log^2 x}$.

For the derivative of the top, rewrite it as $1-e^{-\log\log x/(x+1)}$.

Differentiate. We get $$e^{-\log\log x/(x+1)}\frac{\frac{x+1}{x\log x}-\log\log x }{(x+1)^2} .$$

The ratio of derivative of top to derivative of bottom blows up as $x\to\infty$, so in the long run the terms of our sequence decrease more slowly than the terms of the harmonic series.

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