[For the sake of context:] I am working out a proof that uses Euler's summation formula to show that $H_N-\ln N-\frac{1}{2N}=\gamma+O(1/N^2)$ when $N\to\infty$.
At a certain point in the proof it is said that the estimate $$ \int_0^{1/2}t\left(\frac{1}{(n-t+1/2)^2}-\frac{1}{(n+t+1/2)^2}\right)\,dt\leq \frac{2}{(n-1)^3}\int_0^{1/2}2t^2\,dt\,,\,\,\,n\geq2, $$ is justified by the Mean Value Theorem. I am familiar with the MVT but I wasn't able to understand how it is used here so far. Am I missing something obvious?
Define $f \colon [-\frac 12,\frac 12]\to \mathbf R$ by $$ f(x) = \frac 1{(n+x+\frac 12)^2} $$ Then $f$ is continuously differentiable with $$ f'(x) = -\frac{2}{(n+x+\frac 12)^3} $$ Hence, for each $t\in [0,\frac 12]$, there is some $\tau \in [-t,t]$ such that $$ f(-t) - f(t) = 2t\cdot f'(\tau) $$ Now, as $|\tau|\le \frac 12$, we have $$ f(-t)-f(t) = 2t f'(\tau) \le 2t \frac{2}{(n-\frac 12 - \frac 12)^3} = 2t \cdot \frac{2}{(n-1)^3} $$ This gives your estimate.