Is the series $$\sum_{n=1}^\infty \left(\dfrac {\log n}{\log (n+1)} \right)^{n^2\log n}$$ convergent? I tried to apply some test, or compare with known series, but with no success; it looks too complicated. Please help. Thanks in advance.
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$\begingroup$ where did this series come from? is there a reason you think it might be convergent? $\endgroup$– Dando18Commented Jul 27, 2017 at 15:05
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$\begingroup$ $\left[(1+\frac{\ln(n)-\ln(n+1)}{\ln(n+1)})^{\frac{ln(n+1)}{\ln(n)-\ln(n+1)}}\right]^{\frac{n^2\ln(n)(\ln(n)- ln(n+1))}{\ln(n+1)}}$. Observe the first exponential goes to $e$ and the second exponent goes to $-\infty$ faster than $-n$. $\endgroup$– HellenCommented Jul 27, 2017 at 15:10
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$\begingroup$ It should converge. A quick look at the graph of $a_n$ shows it's way smaller than $\frac{1}{n^{1.1}}$. Now the hard job is playing with inequalities. $\endgroup$– Noé ACCommented Jul 27, 2017 at 15:16
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2 Answers
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From a Taylor expansion to low-order terms: $$\begin{align} \frac{\log n}{\log(n+1)} &= \frac{\log n}{\log n+\log(1+\frac{1}{n})} = \frac{1}{1+\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)} \\&= 1-\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right) \end{align}$$ from which $$\begin{align} \left(\frac{\log n}{\log(n+1)}\right)^{n^2\log n} &=\exp\left(n^2\log n\log\left(\frac{\log n}{\log(n+1)}\right)\right) \\ &=\exp\left(n^2\log n\log\left(1-\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)\right)\right)\\ &=\exp\left(n^2\log n\left(-\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)\right)\right)\\ &=\exp\left(-n+o\left(n\right)\right) =\frac{1}{e^{n+o(n)}} \end{align}$$
Thus, by comparison, the series will converge, since $\sum_n \frac{1}{e^n}$ does.
answered Jul 27, 2017 at 15:14
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$\begingroup$ (Note: by doing a Taylor series expansion to more terms, you can even have the slightly tighter estimate asymptotics $e^{-n+\frac{1}{2}+o(1)}$.) $\endgroup$ Commented Jul 27, 2017 at 15:17
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$\begingroup$ also note the singularity for $n=1$ if you leave $0^0$ undefined. $\endgroup$– Dando18Commented Jul 27, 2017 at 15:24
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Hint. Note that \begin{align*} \left(\dfrac {\log n}{\log (n+1)} \right)^{n^2\log n}&=\left(1+\frac{\log(1+1/n)}{\log(n)}\right)^{-n^2\log n}\\&= \exp\left(-n^2\log(n)\log\left(1+\frac{\log(1+1/n)}{\log(n)}\right)\right)\\ &= \exp\left(-n\cdot n\log(n)\cdot\log\left(1+\frac{1+o(1)}{n\log(n)}\right)\right)\\ &= \exp\left(-n(1+o(1)\right). \end{align*}
answered Jul 27, 2017 at 15:14
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$\begingroup$ I believe I was slightly faster this time :) $\endgroup$ Commented Jul 27, 2017 at 15:15
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1$\begingroup$ @Clement C. Very fast indeed!! (+1) $\endgroup$– Robert ZCommented Jul 27, 2017 at 15:16