Limits Part 1

The key to solve this problem is to realize that the expression (inside limit) is same as integral of this function: integral of (x^2)(1+(x/n))^n dx under the limits 0 to 1.The above equality can be observed easily by use of binomial theorem.The expression (1+(x/n))^n can be rewritten as [(1+(x/n))^(n/x)]^x which, when n tends to infinity becomes e^x.So, the required limit is integral of (x^2)(e^x) dx under the limits 0 to 1.By using integration by parts 2 times, the value of the integral gets evaluated as (e-2).Hence, the value of limit is (e-2).

$\begin{aligned} & \lim_{n\to\infty} \sum_{k=0}^{n} \frac{ \binom {n} {k}}{n^{k}(k+3)} =\lim_{n\to\infty} \int_{0}^{1} x^2(1+\frac{x}{n})^{n} dx \\ &= \int_{0}^{1} x^2 \lim_{n\to\infty} (1+\frac{x}{n})^{n} dx \\ &= \int_{0}^{1} x^2 e^x dx \\ &\text{Integrating by parts} \\ &=x^2\int_{0}^{1} e^x - \int_{0}^{1} (2x)(\int_{0}^{1} e^x dx) dx \\ &=x^2\int_{0}^{1} e^x - (2x)\int_{0}^{1} e^x dx + \int_{0}^{1} (2)(\int_{0}^{1} e^x dx) dx \\ &=e-2 \end{aligned}$ Here we evaluate the limit, $\begin{aligned} &\lim_{n\to\infty} (1+\frac{x}{n})^{n} &&&&&& (1^\infty form) \\ &= e^{\lim_{n\to\infty}\frac{x}{n} \times n} \\ &= e^ {x} \end{aligned}$