We got this question as a bonus for our homework assignment and I'm having trouble figuring out how to start to solve this limit. We need to find $\lim_{n\rightarrow\infty}(a_n)$ for this sequence:
$$a_n=\frac{n+1}{n}\ln(\sqrt{n})-\frac{1}{n^2}\sum^n_{k=1}\ln(k+n)^k$$
Can anyone give suggestions on how to solve this? (any help would be appreciated)
p.s. This was in the 'Riemann sums' section of our homework assignment
We define \begin{align} g(n) = \frac{1}{n^2}\sum^n_{k=1}k\ln\left( k+n \right). \end{align} Since $f(x) = x \ln(x + n)$ is an increasing function, we find \begin{align} n^{-2}\int_{0}^{n}\mathrm{d} x~f(x) < ~& g(n) < n^{-2}\int_{1}^{n+1}\mathrm{d} x~f(x) \\ \frac{1+2\ln n}{4}<~&g(n) < \frac{n^2 - 1}{2n^{2}}\ln(n+1) + \frac{n-2}{4n} + \frac{2n+1}{2n}\frac{\ln(2n+1)}{n}. \end{align} Therefore, we obtain \begin{align} \frac{\ln n}{2n} - \frac{1}{4}< a_{n} < \frac{n+1}{2n}\left(\ln n - \frac{n - 1}{n}\ln(n+1)\right) - \frac{n-2}{4n} + \frac{2n+1}{2n}\frac{\ln(2n+1)}{n}. \end{align} Since both LHS and RHS approach $-1/4$ in the limit of $n \to \infty$, we can conclude $$ \lim_{n \to \infty} a_{n} = -\frac{1}{4}. $$