# On the nature of a series involving ratios and power of logarithm

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.

• where did this series come from? is there a reason you think it might be convergent? – Dando18 Jul 27 '17 at 15:05
• $\left[(1+\frac{\ln(n)-\ln(n+1)}{\ln(n+1)})^{\frac{ln(n+1)}{\ln(n)-\ln(n+1)}}\right]^{\frac{n^2\ln(n)(\ln(n)- ln(n+1))}{\ln(n+1)}}$. Observe the first exponential goes to $e$ and the second exponent goes to $-\infty$ faster than $-n$. – Hellen Jul 27 '17 at 15:10
• 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. – Noé AC Jul 27 '17 at 15:16

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.
• (Note: by doing a Taylor series expansion to more terms, you can even have the slightly tighter estimate asymptotics $e^{-n+\frac{1}{2}+o(1)}$.) – Clement C. Jul 27 '17 at 15:17
• also note the singularity for $n=1$ if you leave $0^0$ undefined. – Dando18 Jul 27 '17 at 15:24