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\problemgrouptitle{Sequences}

\begin{problemblock}[3]{Find the first four terms of the infinite sequence and evaluate $\displaystyle\lim_{n \to \infty}{a_n} $ if it exists.}

\item $\displaystyle a_n=\frac {n^3+3n^2}{2n^3+5}$
\answer{
$\displaystyle a_1=\frac 47, a_2=\frac {20}{21},  a_3=\frac {54}{59},  a_4=\frac{112}{133},   \lim_{n \to \infty}{a_n}=\frac 12$
}

\item  $ \displaystyle a_n=\frac {2 e^n}{3 e^n+1}$
\answer{
$\displaystyle a_1=\frac {2e}{3e+1}, a_2=\frac {2e^2}{3e^2+1},  a_3=\frac {2e^3}{3e^3+1}, a_4=\frac{2e^4}{3e^4+1},   \lim_{n \to \infty}{a_n}=\frac 23$
}

\item  $ \displaystyle a_n=(-1)^n \frac {n^2+3}{n^3+2n+4}$
\answer{
$\displaystyle a_1=-\frac 47, a_2=\frac {7}{16},  a_3=-\frac {12}{37},  a_4=\frac{19}{76},   \lim_{n \to \infty}{a_n}=0$
}

\end{problemblock}

\begin{problemblock}[2]{Evaluate $\displaystyle\lim_{n \to \infty}{a_n} $ if it exists and determine whether the infinite sequence is convergent or divergent.}

\item $\displaystyle a_n=\frac {e^n}{2e^n+n}$
\answer{
$\displaystyle \lim_{n \to \infty}{a_n}=\frac 12$,  convergent 
}

\item $\displaystyle a_n=\frac {\tan^{-1} n}{n}$
\answer{
$\displaystyle \lim_{n \to \infty}{a_n}=0$,  convergent 
}

\item $\displaystyle a_n=\frac {\ln n}{3n}$
\answer{
$\displaystyle \lim_{n \to \infty}{a_n}=0$,  convergent 
}

\item $\displaystyle a_n=(-1)^n\frac {n+1}{n^2+5}$
\answer{
$\displaystyle \lim_{n \to \infty}{a_n}=0$,  convergent 
}

\end{problemblock}

\begin{problemblock}[3]{Determine whether the infinite sequence is increasing, decreasing or not monotonic.}

\item  $\displaystyle a_n= {ne^{-n^2}}$
\answer{
decreasing
}

\item  $\displaystyle a_n=\frac {n^2+3}{n^2+5}$
\answer{
increasing
}

\item $\displaystyle a_n=\frac {e^n}{e^n+2}$
\answer{
increasing
}

\end{problemblock}

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