# How do I prove the following inequality $\sum_{k=n+1}^{\infty}\frac{1}{k^{2}\log k}\leq\frac{1}{n\log n}$?

I would appreciate some help proving the inequality $$\sum_{k=n+1}^{\infty}\frac{1}{k^{2}\log k}\leq\frac{1}{n\log n}.$$ Thanks in advance!

• The function $t \mapsto \frac{1}{t^2\log t}$ is strictly decreasing on $(1,\infty)$, therefore $$\sum_{k=n+1}^\infty \frac{1}{k^2\log k} < \int_n^\infty \frac{dt}{t^2\log t}.$$ – Daniel Fischer Jul 3 '14 at 12:56

The function $x\mapsto \frac{1}{x^2\log x}$ is decreasing so $$\frac{1}{k^2\log k}\le\int_{k-1}^k\frac{dx}{x^2\log x}$$ so $$\sum_{k=n+1}^\infty\frac{1}{k^2\log k}\le\sum_{k=n+1}^\infty\int_{k-1}^k\frac{dx}{x^2\log x}=\int_n^\infty \frac{dx}{x^2\log x}\le\frac1{\log n}\int_n^\infty\frac{dx}{x^2}=\frac{1}{n\log n}$$