Limit of summation

From binomial theorem we have

$\begin{aligned} (1+x)^n & = \sum_{k=0}^n \binom nk x^k & \small \color{#3D99F6} \text{Multiply both sides by}x^2 \\ x^2(1+x)^n & = \sum_{k=0}^n \binom nk x^{k+2} & \small \color{#3D99F6} \text{Integrate both sides from 0 to }\frac 1n \\ \int_0^\frac 1n x^2(1+x)^n dx & = \sum_{k=0}^n \frac {\binom nk}{k+3} x^{k+3} \bigg|_0^\frac 1n \\ & = \frac 1{n^3} \sum_{k=0}^n \frac {\binom nk}{n^k(k+3)} \end{aligned}$

Therefore,

$\begin{aligned} L & = \lim_{n \to \infty} \sum_{k=0}^n \frac {\binom nk}{n^k(k+3)} \\ & = \lim_{n \to \infty} n^3 \color{#3D99F6} \int_0^\frac 1n x^2(1+x)^n dx & \small \color{#3D99F6} \text{By integration by part} \\ & = \lim_{n \to \infty} n^3 \color{#3D99F6} \left[\frac {x^2(1+x)^{n+1}}{n+1}-\frac {2x(1+x)^{n+2}}{(n+1)(n+2)} +\frac {2(1+x)^{n+3}}{(n+1)(n+2)(n+3)} \right]_0^\frac 1n \\ & = \lim_{n \to \infty} \left[\left(1+\frac 1n \right)^n - \frac {2\left(1+\frac 1n\right)^{n+1}}{1+\frac 2n} + \frac {2\left(1+\frac 1n \right)^{n+2}}{\left(1+\frac 2n\right)\left(1+\frac 3n\right)} - \frac 2{\left(1+\frac 1n\right)\left(1+\frac 2n\right)\left(1+\frac 3n\right)} \right] & \small \color{#3D99F6} \text{Note that}\lim_{n \to \infty} \left(1+\frac 1n\right)^n = e \\ & = e - 2e+2e - 2 = e-2 \approx \boxed {0.718} \end{aligned}$

Observe that the summation is

$\sum_{k=0}^n\frac{\operatorname{nCr}\left(n,k\right)}{k+3}\cdot\left(\frac{1}{n}\right)^k$

$=\sum_{k=0}^n\int_0^1x^{\left(k+2\right)}\operatorname{nCr}\left(n,k\right)\cdot\left(\frac{1}{n}\right)^kdx$

$=\int_0^1\left(x^2\sum_{k=0}^n\operatorname{nCr}\left(n,k\right)\cdot\left(\frac{x}{n}\right)^k\right)dx$

$=\int_0^1x^2\left(1+\frac{x}{n}\right)^ndx$

Now, letting $\displaystyle n$ tend to $\displaystyle \infty$ the last integral turns to

$\int_0^1x^2e^xdx$

This can be easily evaluated and thus the answer comes out to be $\displaystyle \boxed {e-2}$ .

Additional Content:

Note that using the same technique we can generalise the sum given in the question. For any real number $\displaystyle m\ge1$

$\sum_{k=0}^n\frac{\operatorname{nCr}\left(n,k\right)}{k+m}\cdot\left(\frac{1}{n}\right)^k=\int_0^1x^{\left(m-1\right)}e^xdx$

when $\displaystyle n$ tends to $\displaystyle \infty$ .