# Riemann sum $\lim_{n\to \infty}\frac2n\sum_{k=1}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)$ to integral

I don't know how to make the following limit $$\displaystyle \lim_{n\to \infty}\frac2n\sum_{k=1}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)$$

into a definite integral and just need some guiding help anything will help thanks.

• What is your problem? Hint: $\int_0^1 f(x)\,dx = \lim_{n\to\infty}\tfrac 1 n\sum_{k=1}^nf(\tfrac k n)$. – amsmath Sep 8 '18 at 22:37

$$\lim_{n\to \infty}\frac1n\sum_{k=0}^{n} f\left(a+{k\over n}(b-a)\right)=\int_a^b f(x) dx$$
$$\lim_{n\to \infty}\frac2n\sum_{k=0}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)=2\int_0^1(2+x)\ln(2+x)\,dx$$