Evaluate $\lim\limits_{n\rightarrow \infty} \sum\limits_{r=0}^{\infty} \frac{^n\text{C}_{r}}{n^r(r+4)}$

Question

$$\displaystyle \lim_{n\rightarrow \infty} \sum_{r=0}^{\infty} \frac{^n\text{C}_{r}}{n^r(r+4)}$$

The answer is $6-2e$ according to WolframAlpha: Link

I first thought it was the expansion of $(1+\frac{1}{n})^n$, but that r+4 in the denominator is causing problems

Answer 1

Method 1. Notice that for each fixed $k \geq 0$, we have

$$ \frac{\binom{n}{k}}{n^k (k+4)} = \frac{n(n-1)\cdots(n-k+1)}{n^k k! (k+4)} = \frac{1}{k! (k+4)} \prod_{j=1}^{k} \left( 1 - \frac{j-1}{n} \right). $$

Now this increases in $n$ and converges to $\frac{1}{k!(k+4)}$ as $n \to \infty$. So by the monotone convergence theorem,

$$ \lim_{n\to\infty} \sum_{k=0}^{\infty} \frac{\binom{n}{k}}{n^k (k+4)} = \sum_{k=0}^{\infty} \lim_{n\to\infty} \frac{\binom{n}{k}}{n^k (k+4)} = \sum_{k=0}^{\infty} \frac{1}{k!(k+4)}. $$

The last sum can be computed by writing

$$\sum_{k=0}^{\infty} \frac{1}{k!(k+4)} = \int_{0}^{1} x^3 e^x \, dx = \left[ (x^3 - 3x^2 + 6x - 6) e^x \right]_{0}^{1} = 6 - 2e. $$

Method 2. As in other answers, you find that

$$ \sum_{k=0}^{\infty} \frac{\binom{n}{k}}{n^k (k+4)} = n^4 \int_{0}^{\frac{1}{n}} t^3 (1 + t)^n \, dt \stackrel{nt = u}{=} \int_{0}^{1} u^3 \left(1 + \frac{u}{n} \right)^n \, du. $$

Taking $n\to\infty$, it follows that $\left(1 + \frac{u}{n} \right)^n \to e^u$ and hence by the dominated convergence theorem (or by the monotone convergence theorem, upon proving that the integrands increase in $n$),

$$ \lim_{n\to\infty} \sum_{k=0}^{\infty} \frac{\binom{n}{k}}{n^k (k+4)} = \int_{0}^{1} u^3 e^u \, du = 6 - 2e. $$

Alternatively, we can utilize the fact that the integrands converge uniformly, which can be proved either by brutal force or by Dini's theorem.

Answer 2

Hints :

First show that if $S_n$ is your sum, then $$S_n=n^4\int_0^{1/n}t^3(1+t)^n dt$$ and compute the integral, using tfat you can find constants $a_0,a_1,a_2,a_3$ such that $$t^3=a_0+a_1(t+1)+a_2(t+1)^2+a_3(t+1)^3$$

Answer 3

First of all you need to compute:

$$S_n=\sum_{r=0}^{\infty} \frac{^n\text{C}_{r}}{n^r(r+4)}.$$

This can be done in usual way: $$ S_n=n^4\int_0^{1/n} dx \sum_{r=0}^{\infty}\binom{n}{r}x^{r+3}= n^4\int_0^{1/n} x^3(1+x)^n\,dx\\ =n^4\left[ \frac{x^3(1+x)^{n+1}}{n+1}-\frac{3x^2(1+x)^{n+2}}{(n+1)(n+2)} +\frac{6x(1+x)^{n+3}}{(n+1)(n+2)(n+3)}-\frac{6(1+x)^{n+4}}{(n+1)(n+2)(n+3)(n+4)}\right]_0^{1/n}\\ =\frac{n(1+\frac{1}{n})^{n+1}}{n+1}-\frac{3n^2(1+\frac{1}{n})^{n+2}}{(n+1)(n+2)} +\frac{6n^3(1+\frac{1}{n})^{n+3}}{(n+1)(n+2)(n+3)}-\frac{6n^4(1+\frac{1}{n})^{n+4}-6n^4}{(n+1)(n+2)(n+3)(n+4)}, $$ so that $$ \lim_{n->\infty}S_n=e-3e+6e-(6e-6)=6-2e. $$