The tails of sums of reciprocal powers have nice estimates: For $\alpha>1$ the integral test gives $$ \sum_{n=j}^\infty \frac{1}{n^\alpha} \leq \int_{j-1}^\infty \frac{1}{x^\alpha} dx = \frac{1}{\alpha-1}\frac{1}{(j-1)^{\alpha-1}} $$ i.e. the tail behaves like $O(1/j^{\alpha-1})$.
The series $$ \sum \frac{1}{(n+\ln(n))^\alpha} $$ converges a bit faster than $\sum 1/n^\alpha$, and hence, the tail $$ \sum_{n=j}^\infty \frac{1}{(n+\ln(n))^\alpha} $$ should behave better than $O(1/j^{\alpha-1})$ and I am interested in how fast does the tail goes to zero. In other words:
What is the asymptotics of the tail $\sum_{n=j}^\infty \frac{1}{(n+\ln(n))^\alpha}$?
(In fact, I am most interested in $\alpha=3$…)
Since $0\le\ln n\le n$, we have $$ \frac{1}{2^\alpha\,n^\alpha}\le\frac{1}{(n+\ln n)^\alpha}\le\frac{1}{n^\alpha}. $$ The tail of $\sum(n+\ln n)^{-\alpha}$ is thus of the same order as the tail of $\sum n^{-\alpha}$.