
Problem 4. a) Let $n \geq 0$ be an integer. Calculate $\int_{0}^{1}(1-t)^{n} e^{t} d t$.

b) Let $k \geq 0$ be a fixed integer and let $\left(x_{n}\right)_{n \geq k}$ be the sequence defined by

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

Prove that the sequence converges and find its limit.


Solution. a) Let $I_{n}=\int_{0}^{1}(1-t)^{n} e^{t} d t, n \geq 0$. We integrate by parts and we get that $I_{n}=-1+n I_{n-1}$, $n \geq 1$ which implies that $\frac{I_{n}}{n!}=-\frac{1}{n!}+\frac{I_{n-1}}{(n-1)!}$. It follows that

$$
\frac{I_{n}}{n!}=I_{0}-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{n!}=e-1-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{n!}
$$

Thus,

$$
I_{n}=n!\left(e-1-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{n!}\right), n \geq 0
$$

b) We have

$$
x_{n+1}-x_{n}=\binom{n+1}{k}\left(e-1-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{(n+1)!}\right)>0
$$

hence the sequence is strictly increasing.

On the other hand, based on Taylor's formula, we have that

$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}+\frac{e^{\theta}}{(n+1)!}
$$

for some $\theta \in(0,1)$. It follows that

$$
0<e-1-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{n!}<\frac{e}{(n+1)!}
$$

Therefore

$$
x_{n} \leq \sum_{i=k}^{n}\binom{i}{k} \frac{e}{(i+1)!} \leq \frac{e}{k!} \sum_{i=k}^{n} \frac{1}{(i-k)!}=\frac{e}{k!}\left(\frac{1}{0!}+\frac{1}{1!}+\ldots+\frac{1}{(n-k)!}\right) \leq \frac{e^{2}}{k!}
$$

which implies the sequence is bounded. Since the sequence is bounded and increasing it converges.

To find $\lim _{n \rightarrow \infty} x_{n}$ we apply part a) of the problem and we have, since

$$
e-1-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{i!}=\frac{1}{i!} \int_{0}^{1}(1-t)^{i} e^{t} d t
$$

that

$$
x_{n}=\sum_{i=k}^{n}\binom{i}{k} \frac{1}{i!} \int_{0}^{1}(1-t)^{i} e^{t} d t=\frac{1}{k!} \int_{0}^{1}(1-t)^{k} e^{t}\left(\sum_{i=k}^{n} \frac{(1-t)^{i-k}}{(i-k)!}\right) d t
$$

Since $\lim _{n \rightarrow \infty} \sum_{i=k}^{n} \frac{(1-t)^{i-k}}{(i-k)!}=e^{1-t}$ and $\sum_{i=k}^{n} \frac{(1-t)^{i-k}}{(i-k)!}<e^{1-t}$, we get based on Lebesgue Dominated Convergence Theorem

$$
\lim _{n \rightarrow \infty} x_{n}=\frac{1}{k!} \int_{0}^{1}(1-t)^{k} e^{t} e^{1-t} d t=\frac{e}{(k+1)!}
$$

Remark. Part b) of the problem has an equivalent formulation

$$
\sum_{i=k}^{\infty}\binom{i}{k}\left(e-1-\frac{1}{1!}-\frac{1}{2!}-\ldots-\frac{1}{i!}\right)=\frac{e}{(k+1)!}
$$

