## Subject IV. (20 points)

Consider the sequences $\left(a_{n}\right)_{n>1}$ and $\left(b_{n}\right)_{n>1}$, where: $a_{n}=\sum_{k=1}^{n} \frac{k+2}{k!+(k+1)!+(k+2)!}$ and $b_{n}=\sum_{k=1}^{n} k!(k+2)$

Required:
a) $\lim _{n \rightarrow \infty} a_{n}$;
b) $\lim _{n \rightarrow \infty}\left(2 a_{n}\right)^{b_{n}}$.

Prof. Alb Nicolae, Lic. T. “O. Goga” Huedin

All questions are mandatory. 10 points are awarded by default.

Time allowed for work - 3 hours.

## Grading Scale for 11th Grade

(OLM 2013 - local stage)

## Default: 10 points
## Subject IV

a) $a_{n}=\sum_{k=1}^{n} \frac{k+2}{k![1+k+1+(k+1)(k+2)]}=\sum_{k=1}^{n} \frac{1}{k!(k+2)}=\sum_{k=1}^{n} \frac{1}{k!(k+2)}=\sum_{k=1}^{n}\left[\frac{1}{(k+1)!}-\frac{1}{(k+2)!}\right]=\frac{1}{2}-\frac{1}{(n+2)!} \lim _{n \rightarrow \infty} a_{n}=\frac{1}{2}$.

(10 points)

b) Clearly $\lim _{n \rightarrow \infty} b_{n}=+\infty$ and thus the limit we are asked to find is in the case $1^{\infty}$.

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(2 a_{n}\right)^{b_{n}}=\lim _{n \rightarrow \infty}\left[1-\frac{2}{(n+2)!}\right]^{b_{n}}=e^{\lim _{n \rightarrow \infty}\left[\frac{2}{(n+2)!} \cdot b_{n}\right]} . \\
& \lim _{n \rightarrow \infty} \frac{b_{n}}{(n+2)!}=\lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n} k!(k+2)}{(n+2)!} \stackrel{\text { Stolz } C}{=} \lim _{n \rightarrow \infty} \frac{(n+1)!(n+3)}{(n+3)!-(n+2)!}=\lim _{n \rightarrow \infty} \frac{(n+1)!(n+3)}{(n+2)!(n+2)}=\lim _{n \rightarrow \infty} \frac{(n+3)}{(n+2)^{2}}=0 .
\end{aligned}
$$

Thus: $\lim _{n \rightarrow \infty}\left(2 a_{n}\right)^{b_{n}}=e^{0}=1$

(10 points)