## Subject III. (40 points)

a) Let $\left(x_{n}\right)_{n \geq 1}$ be a sequence of real numbers defined by the formula $x_{n}=\frac{\frac{1}{2015}+\frac{1}{2016}+\ldots+\frac{1}{n+2014}}{\ln (n+2015)}$.

Calculate $\lim _{n \rightarrow \infty} x_{n}$;

Prof. Raul Domsa, Liceul Teoretic „Petru Maior” Gherla

b) Consider the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$, where $a_{n}=\frac{1}{1 \cdot 3}-\frac{1}{2 \cdot 4}+\frac{1}{3 \cdot 5}-\frac{1}{4 \cdot 6}+\ldots+\frac{1}{(2 n-1)(2 n+1)}, n \geq 1$ and $b_{n}=1 \cdot 3-2 \cdot 4+3 \cdot 5-4 \cdot 6+\ldots+(2 n-1)(2 n+1)$. Calculate $\lim _{n \rightarrow \infty}\left(4 a_{n}\right)^{b_{n}}$.

Prof. Nicolae Alb, Liceul Teoretic „Octavian Goga” Huedin
## Subject III. (40 points)

a) We use the Cesaro-Stolz criterion for \(a_{n}=\frac{1}{2015}+\frac{1}{2016}+\ldots+\frac{1}{n+2014}\) and \(b_{n}=\ln (n+2015), b_{n} \rightarrow \infty\), increasing.

\[
\lim _{n \rightarrow \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n+2015}}{\ln \left(\frac{n+2016}{n+2015}\right)}=\lim _{n \rightarrow \infty} \frac{1}{\ln \left(1+\frac{1}{n+2015}\right)^{n+2015}}=\frac{1}{\ln \left(\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n+2015}\right)^{n+2015}\right]\right.}=\frac{1}{\ln e}=1
\]

\[
\operatorname{From}(*),(* *) \stackrel{\text { c.C.-S. }}{\Rightarrow} \lim _{n \rightarrow \infty} x_{n}=1;
\]

(20 p)

b) \(a_{n}=\frac{1}{4} \cdot \frac{2 n^{2}+n+1}{2 n^{2}+n}\)

(10 p) \(\quad b_{n}=2 n^{2}+n\)

(5 p)

\[
\lim _{n \rightarrow \infty}\left(4 a_{n}\right)^{b_{n}}=e
\]