Question

1. $ \left\{\frac{2 n^{2}}{n^{2}+1}\right\}, a_{1}=1, a_{2}=\frac{8}{5}, a_{3}=1,8, a_{4}=\frac{32}{17}=1 \frac{15}{17} \ldots $ \[ \frac{2 n^{2}+2-2}{n^{2}+1}=2-\frac{2}{n^{2}+1} \Varangle \rightarrow 2 \] a) bounden above; $ \quad\left\{a_{n}\right\} \rightarrow 2 $ b) positive c) increasing $ \left(-a_{n}\right)+a_{n+1}=\frac{2 n^{2}+4 n+2}{n^{2}+2 n+2}-\frac{2 n^{2}}{n^{2}+1}=\frac{2 n^{4}+2 n^{2}+4_{n}^{3}-4 n+2 n^{2}+2}{\left(n^{2}+1\right)\left(n^{2}+2 n^{2}+2\right)}- $ \[ \begin{aligned} -\frac{2 n^{4}+4{ }^{3}+4{ }^{2}}{\left(n^{2}+1\right)\left(n^{2}+2 n+2\right)} & =\frac{2}{\frac{\left(n^{2}+1\right)\left((n+1)^{2}+1\right)}{70}}>0 \\ a_{n+1} & >a_{n} \end{aligned} \] $ \left\{a_{n}\right\} $ is increasing d) covergent to 2

          1. \( \left\{\frac{2 n^{2}}{n^{2}+1}\right\}, a_{1}=1, a_{2}=\frac{8}{5}, a_{3}=1,8, a_{4}=\frac{32}{17}=1 \frac{15}{17} \ldots \)
\[
\frac{2 n^{2}+2-2}{n^{2}+1}=2-\frac{2}{n^{2}+1} \Varangle \rightarrow 2
\]
a) bounden above; \( \quad\left\{a_{n}\right\} \rightarrow 2 \)
b) positive
c) increasing \( \left(-a_{n}\right)+a_{n+1}=\frac{2 n^{2}+4 n+2}{n^{2}+2 n+2}-\frac{2 n^{2}}{n^{2}+1}=\frac{2 n^{4}+2 n^{2}+4_{n}^{3}-4 n+2 n^{2}+2}{\left(n^{2}+1\right)\left(n^{2}+2 n^{2}+2\right)}- \)
\[
\begin{aligned}
-\frac{2 n^{4}+4{ }^{3}+4{ }^{2}}{\left(n^{2}+1\right)\left(n^{2}+2 n+2\right)} & =\frac{2}{\frac{\left(n^{2}+1\right)\left((n+1)^{2}+1\right)}{70}}>0 \\
a_{n+1} & >a_{n}
\end{aligned}
\]
\( \left\{a_{n}\right\} \) is increasing
d) covergent to 2

$1. {(2 n^2)/(n^2+1)}, a1=1, a2=(8)/(5), a3=1,8, a4=(32)/(17)=1 (15)/(17)… (2 n^2+2-2)/(n^2+1)=2-(2)/(n^2+1)→ 2 a) bounden above; {an}→ 2 b) positive c) increasing (-an)+an+1=(2 n^2+4 n+2)/(n^2+2 n+2)-(2 n^2)/(n^2+1)=(2 n^4+2 n^2+4n^3-4 n+2 n^2+2)/((n^2+1)(n^2+2 n^2+2))- -(2 n^4+4^3+4^2)/((n^2+1)(n^2+2 n+2)) =(2)/(((n^2+1)((n+1)^2+1))/(70))>0 an+1 >an {an} is increasing d) covergent to 2$

Added by Nathaniel P.

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