Show $\sum_{k=1}^n \frac{1}{k^2} \le 2$ and $\ln(n!) \ge 1 -n+n\ln(n)$ for all positive integers n
For the first inequality: \begin{align*} \sum_{k=1}^{n}\dfrac{1}{k^{2}}&\leq 1+\sum_{k=2}^{n}\dfrac{1}{k(k-1)}\\ &=1+\sum_{k=2}^{n}\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right)\\ &=1+1-\dfrac{1}{n}\\ &<2. \end{align*}
By looking at the upper sum, we see that \begin{align*} \log(n!)&=\sum_{k=1}^{n}\log k\\ &\geq\int_{1}^{n}\log udu\\ &=u\log u\bigg|_{u=1}^{u=n}-\int_{1}^{n}1du\\ &=n\log n-n+1. \end{align*}