Seemous 2017/4

(a) Let $n \in \mathbb{N} \cup \{ 0 \}$ . Evaluate the integral:

$\mathcal{J} = \int_0^1 \left( 1 - t \right)^n e^t \, {\rm d}t$

(b) Let $k \in \mathbb{N} \cup \{ 0 \}$ and let $\{x_n\}_{n \geq k}$ be a sequence defined as

$x_n = \sum_{i=k}^n \binom{i}{k} \left(e - 1 - \frac{1}{1!} - \frac{1}{2!} - \cdots -\frac{1}{i!} \right)$

Find the limit of $x_n$ .

Solution

(a) We have successively:

$\begin{align*} \int_{0}^{1} \left ( 1-t \right )^n e^t \, {\rm d}t &\overset{u=1-t}{=\! =\! =\! =\!} \int_{0}^{1} u^n e^{1-u} \, {\rm d}u\\ &= e \int_{0}^{1}u^n e^{-u} \, {\rm d}u\\ &= e \gamma(n+1, 1)\\ &\!\overset{(*)}{=} e n!\left ( 1- e^{-1} \sum_{k=0}^{n} \frac{1}{k!} \right )\\ &= n! \left ( e - \sum_{k=0}^{n} \frac{1}{k!} \right ) \end{align*}$

where $\gamma$ denotes the lower incomplete Gamma function.

(b) We are invoking Tonelli’s theorem. Hence the limit we seek is equal to

$\begin{align*} \sum_{i=k}^{\infty} \binom{i}{k} \sum_{r=i}^{\infty} \frac{1}{(r+1)!} &= \sum_{r=k}^{\infty} \frac{1}{(r+1)!}\sum_{i=k}^{r} \binom{i}{k} \\ &= \sum_{r=k}^{\infty} \frac{1}{(r+1)!} \binom{r+1}{k+1}\\ &= \frac{1}{\left ( k+1 \right )!} \sum_{r=k}^{\infty} \frac{1}{\left ( r-k \right )!}\\ &= \frac{e}{\left ( k+1 \right )!} \end{align*}$

Seemous 2017/4

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