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

Hint :

$\mathbf{\text{Method 1:}}$

Use that for large enough $n$ we have $$\binom{n}{r}\sim\frac{n^r}{r!}$$

The rest is easy assuming you know the Taylor series of $e$

$\mathbf{\text{Method 2:}}$

Using binomial theorem we have that $$(1+x)^n=\sum_{r=0}^n \binom{n}{r}x^r$$ thus $$x^2(1+x)^n=\sum_{r=0}^n \binom{n}{r}x^{r+2}$$

What happens if you integrate the above equation wrt $x$ from $0$ to $\displaystyle \frac{1}{n}$? (Note that left hand side can be easily integrated using Integration by Parts)

Partial answer:

Using the inequalities $$\frac{1}{r^r} \le n^{-r}\binom{n}{r} \le \frac{1}{r!}$$ we obtain $$\sum_{r=0}^n \frac{1}{r^r(r+3)} \le \sum_{r=0}^n \frac{\binom{n}{r}}{n^r (r+3)} \le \sum_{r=0}^n \frac{1}{r! (r+3)}. \tag{$*$}$$

The right-hand side of ($*$) can be written as \begin{align} \sum_{r=0}^n \frac{1}{r! (r+3)} &= \sum_{r=0}^n \frac{(r+1)(r+2)}{(r+3)!} \\ &= \sum_{r=0}^n \frac{(r+3)(r+2) - 2(r+3) + 2}{(r+3)!} \\ &\overset{n \to \infty}{\longrightarrow} (e - 1) - 2(e - 1 - 1) + 2(e - 1 - 1 - \frac{1}{2}) = e - 2. \end{align}

It remains to lower bound the desired series by $e-2$. Evidently the left-hand side of ($*$) tends to $\approx 0.640 < 0.718 \approx e-2$, so something better is required.