# Test the convergence of $\sum 1/(\log n)^{\log n}.$ [duplicate]

I can not understand that how it is proved, so please somebody help me.I approch by logarithimic test but I unable to find out the convergency or divergency.

$$\sum_{n=2}^\infty\frac1{(\log(n))^{\log(n)}}<\sum_{n=1}^\infty\frac{2^n}{(\log(2^n))^{\log(2^n)}}=\sum_{n=1}^\infty\frac{2^n}{n^{n\log(2)}(\log(2))^{n\log(2)}}<\sum_{n=1}^\infty\frac{2^n}{n^n}$$

And that last sum converges by ratio test, hence your series converges.

• Cauchy condensation is honestly the best when it comes to logarithms! – Simply Beautiful Art Feb 14 '17 at 2:47
• Ok no problem, simply beatiful art – Saroj Ghosh Feb 14 '17 at 3:14

Let $n> e^{e^2}.$ Then $\ln (\ln n) > 2.$ Hence

$$\ln (\ln n)^{\ln n} = (\ln n)\ln (\ln n) > (\ln n)\cdot 2 = \ln n^2.$$

Exponentiating then gives $(\ln n)^{\ln n} > n^2,$ which implies

$$\frac{1}{(\ln n)^{\ln n}} <\frac{1}{n^2}.$$

Since $\sum 1/n^2<\infty,$ the series in question converges by the comparison test.