Find : $\lim_{n\to\infty}\sum_{k=0}^n\frac{\binom{n}{k}}{n^k(k+1)}$

Question

I'm try to find this lim $\lim_{n\to\infty}\sum_{k=0}^n\frac{\binom{n}{k}}{n^{k}(k+1)}$ Is this limits can be done by integral !? Or inequality Someone help me hints me Thanks!

Answer 1

Hint:

Recall that $$(x+1)^n=\sum_{k=0}^{n}{n\choose k} x^k$$ so $$\frac1x\int_0^x(t+1)^ndt=\frac{(x+1)^{n+1}-1}{x(n+1)}=\sum_{k=0}^{n}\frac{{n\choose k}x^k}{k+1}$$ so $$\sum_{k=0}^{n}\frac{{n\choose k}}{n^k(k+1)}=\frac{(\frac1n+1)^{n+1}-1}{\frac1n(n+1)}$$ thus your limit is $$\lim_{n\to \infty}\frac{(\frac1n+1)^{n+1}-1}{\frac1n(n+1)}$$