Evaluate $\lim_{n\to\infty}\sum_{k=0}^n\frac{^nC_k}{n^k(k+3)}$ The following question is taken from JEE practice set.

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

I found its solution on sarthaks website.
They wrote $\frac1{k+3}$ as $\int_0^1x^{k+2}dx$ and then applied binomial theorem.
I understood this method but I just think its first step is not very intuitive as I am coming across this approach for the first time.
Is there any other way to solve it?
Or are there any other similar questions out there?
 A: Let
$$ f_n(x)=\sum_{k=0}^n\frac{^nC_k}{n^k(k+3)}x^{k+3}. $$
Then
$$ f'(x)=\sum_{k=0}^n \frac{^nC_k}{n^k}x^{k+2}=x^2\sum_{k=0}^n {^nC_k}\left(\frac xn\right)^k=x^2\left(1+\frac xn\right)^n $$
and hence
$$ f_n(1)=\int_0^1x^2\left(1+\frac xn\right)^ndx. $$
Since
$$ \left(1+\frac xn\right)^n\le \left(1+\frac 1n\right)^n$$
and
$$ \lim_{n\to\infty}\left(1+\frac 1n\right)^n=e $$
one has, by DCT,
$$ \lim_{n\to\infty}\sum_{k=0}^n\frac{^nC_k}{n^k(k+3)}=\lim_{n\to\infty}f_n(1)=\lim_{n\to\infty}\int_0^1x^2\left(1+\frac xn\right)^ndx=\int_0^1x^2e^xdx. $$
A: The "trick" used is maybe the way to get it written as fluently as possible. An "other" way of thinking about the problem, in smaller steps would be as follows. We have to compute a limit. Its a limit over a complicated sum. So let us compute that sum first. The sum is
$$
S_n :=\sum_{0\le k\le n}\binom nk\frac 1{n^k(k+3)}\ .
$$
(The binomial coefficient is $\binom nk$, using an alternative notation instead of the $C$-notation.)
Now there is a "usual" trick to compute such sums, use a variable $x$, associate the more general sum
$$
S_n(x) :=\sum_{0\le k\le n}\binom nk\frac 1{n^k(k+3)}\cdot x^{k+3}\ ,
$$
and observe that $S_n$ is $S_n(1)$ (sic!), and we know $S_n(0)=0$, and can compute the derivative $S_n'(x)$:
$$
S_n'(x)
=
\sum_{0\le k\le n}\binom nk\frac 1{n^k(k+3)}\cdot (k+3)x^{k+2}
=
\sum_{0\le k\le n}\binom nk\cdot \left(\frac xn\right)^{k+2}
=
x^2\left(1+\frac xn\right)^n\ .
$$
This gives in a relatively natural manner
$$
S_n=S_n(1)-S_n(0)=\int_0^1 S_n'(x) =
\int_0^1
x^2\left(1+\frac xn\right)^n\; dx
\ .
$$
Now take the limit using dominated convergence with dominating function $x^2\exp x$, so that
$$
\lim_n S_0 
=\lim_n
\int_0^1
x^2\left(1+\frac xn\right)^n\; dx
=\int_0^1
\lim_n
x^2\left(1+\frac xn\right)^n\; dx
=\int_0^1
x^2\exp x\; dx
=e-2
\ .
$$

Well, this is in essence the same as the linked solution...
A: Just for your curiosity.
After @xpaul's and @dan_fulea's answers, we can even go beyond the limit considering
$$I_n=\int x^2\left(1+\frac xn\right)^n\; dx $$ Integration by parts leads to
$$I_n=n\,\left(1+\frac xn\right)^{n+1}\,\,\,\frac{2x^2+nx(3x-2)+n^2(x^2-2x+2) }{(n+1)(n+2)(n+3)}$$
$$J_n=\int_0^1 x^2\left(1+\frac xn\right)^n\; dx=\left(1+\frac{1}{n}\right)^n\frac{ \left(n^2+n+2\right)}{(n+2)
   (n+3)}- \frac{2 n^3}{(n+1) (n+2) (n+3)}$$ which makes the limit $(e-2)$ quite clear.
Now, expanding $J_n$ as a series for large values of $n$ leads to
$$J_n=(e-2)-\frac{3 (3 e-8)}{2 n}+\frac{443 e-1200}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
The above truncated series gives for example
$$J_{10}=\frac{1763 e}{2400}-\frac{13}{10}=0.696805$$ while the exact value is
$$J_{10}=\frac{249060564759}{357500000000}=0.696673$$ then a relative error of $0.019$%.
A: Using Binomial theorem expands $$
(1+t)^n=\sum_{k=0}^n n C_k t^k
$$
Multiplying the equation by $t^2$ and integrating from $t=0$ to $t=\frac{1}{n}$ gives
$$
\begin{aligned}
\int_0^{\frac{1}{n}} t^2(1+t)^n d t &=\sum_{k=0}^n n C_k \int_0^{\frac{1}{n}} t^{k+2} d t \\
&=\sum_{k=0}^n \frac{n C_k}{n^{k+3}(k+3)}
\end{aligned}
$$
Hence $$
\boxed{\lim _{n \rightarrow \infty} \sum_{k=0}^n \frac{n C_k}{n^{k}(k+3)}=\lim _{n \rightarrow \infty}\left[n^3\int_0^{\frac{1}{n}} t^2(1+t)^n d t\right]=e-2}
$$
