inequality involving logarithms and summations

Question

Is is true that $\sum \limits_{j=a}^{b} \frac{1}{j^2 \ln^2 j} \leq \frac{1}{(a-1)\ln (a-1)}-\frac{1}{b \ln b}$

for all choices of $a$ and $b$ ?

Answer 1

It is enough to consider a right-hand Riemann sum for the function $f(x) = 1/(x\ln(x))^2$ on the interval $(a-1, b)$ and note that, when $a-1>1$, it follows that $$ \frac{1}{(x\ln(x))^2} < \frac{1+\ln(x)}{(x\ln(x))^2} $$ for $a-1<x<b$. Integrating the function on the right-hand side of the above inequality yields the right-hand side of the inequality you wish to establish. I'll leave the rest (including valid values for $a$) to you.

Answer 2

Hint: For integer $b\geqslant j\geqslant a> 2$, show

$$\frac1{(j \log j)^2} < \frac1{j \log j \cdot (j-1)\log (j-1)} < \frac1{(j-1)\log(j-1)} - \frac1{j \log j} $$

Now sum telescopically...