If $\lim_{n \to \infty} \sum^n_{k =0} \frac{nC_k}{n^k(k+3)} =ae+b$

Question

Problem :

If $$\lim_{n \to \infty} \sum^n_{k =0} \frac{nC_k}{n^k(k+3)} =ae+b$$

then find the value of a-b.

Please suggest how to solve such problem in whicl summation and limits both are there. Thanks.

Answer 1

Note that:

$ x^2\bigg(1+\dfrac{x}{n}\bigg)^n = \displaystyle \sum_{k=0}^{n} {n \choose k} \dfrac{x^{k+2}}{n^k} $

Integrating both sides with limits 0 to 1,

$\displaystyle \sum_{k=0}^{n} \dfrac{{n \choose k}}{n^k (k+3)} = \int_{0}^{1} x^2\bigg(1+\dfrac{x}{n}\bigg)^n {\mathrm dx} $

Now , $ \displaystyle \lim_{n \to \infty} \bigg(1+\dfrac{x}{n}\bigg)^n = e^x $,

Hence, $\displaystyle \lim_{n \to \infty} \displaystyle \sum_{k=0}^{n} {n \choose k} \dfrac{x^{k+2}}{n^k} = \int_{0}^{1} x^2 e^x {\mathrm dx} $

$= e - 2 $

Hence, $a - b = 3$