$ \lim_{n\to \infty} \sum_{k=0}^n \frac {\binom{n}{k}}{(n^k)(k+3)}$

Question

How to find this limit? I am not able to convert it into the right form so that I can integrate it ? Please help me out. Thank you.

Answer 1

$$ f(x)=\sum_{k=0}^{n} {\binom{n}{k}x^k} = (1+x)^n$$

Now, the given sum is $$\sum_{k=0}^{n} {\binom{n}{k}}\frac{x^k}{k+3}$$

$=\frac{1}{x³}\int_{0}^{x} t²f(t) dt = \frac{1}{x³}\int_{0}^{x} t²(1+t)^n dt$

Because, $$\frac{1}{x³}\int_{0}^{x} t²f(t) dt =\frac{1}{x²}\int_{0}^{x} \left(\sum_{k=0}^{n} {\binom{n}{k}} t^{k+2} \right) dt =\frac{1}{x³}\left(\sum_{k=0}^{n} \binom{n}{k}\frac{x^{k+3}}{k+3}\right)$$

Now,

$=\frac{1}{x³}\int_{0}^{x} t²f(t) dt = \frac{1}{x³}\int_{0}^{x} t²(1+t)^n dt$

$ =\frac{1}{x³}[\frac{(1+x)^{(1+n)}((n²+3n+2)x²-2(n+1)x+2)}{(n+1)(n+2)(n+3)} -\frac{2}{(n+1)(n+2)(n+3)}] $.

Now, putting $x=\frac{1}{n}$ and letting $n \to \infty$ we have to solve

$\lim \limits_{n \to \infty} n^3[\frac{(1+1/n)^{(1+n)}((n²+3n+2)(1/n)²-2(1+1/n)x+2)}{(n+1)(n+2)(n+3)}] \\- \lim \limits_{n \to \infty} n³[\frac{2}{(n+1)(n+2)(n+3)}] $.

$ =\lim \limits_{n \to \infty}\\ (1+1/n)^n((n²+3n+2)(1/n²)-2(1+1/n)+2) -2$

$=\lim \limits_{n \to \infty} e(1+3/n+2/n² -2+2) +2 =e-2$