Big Theta $\sum_{i=1}^n i^2\ln(e+\frac{1}{i})=\Theta(n^3)$

Question

Please help explain $\sum_{i=1}^n i^2\ln(e+\frac{1}{i})=\Theta(n^3)$. I intuitively get it, but would appreciate if someone could actually provide a rigorous approach.

Answer 1

Hint: use that $1 \lt \ln \left(e+\frac{1}{n}\right) \lt 2$, for $n \gt 1$.