Question-36
Consider a sequence \(\displaystyle \{a_{n}\}\) with \(\displaystyle a_{n} =\frac{n^{2} +1}{n^{2}}\). Let \(\displaystyle \{b_{n}\}\) be another sequence defined as \(\displaystyle b_{n} =2a_{n}^{2} +3a_{n} +1\). Which of the following options are true?
We have:
\[ \begin{equation*} a_{n} =\frac{n^{2} +1}{n^{2}} =1+\frac{1}{n^{2}} \end{equation*} \]
As \(\displaystyle n\uparrow\), \(\displaystyle \frac{1}{n^{2}} \downarrow\), as a result, \(\displaystyle a_{n} \downarrow\). We also see that \(\displaystyle a_{n}\) is bounded since \(\displaystyle \lim\limits _{n\rightarrow \infty } \ a_{n} =1\) and the sequence is decreasing. We have \(\displaystyle a_{n} \leqslant 2\) for all \(\displaystyle n\in \mathbb{N}\).
We have:
\[ \begin{equation*} \begin{aligned} b_{n} & =2a_{n}^{2} +3a_{n} +1 \end{aligned} \end{equation*} \]
We see that \(\displaystyle b_{n}\) is also decreasing since it is a polynomial in \(\displaystyle a_{n}\) with positive coefficients. Using algebra of limits:
\[ \begin{equation*} \lim\limits _{n\rightarrow \infty } \ b_{n} =2\times 1+3\times 1+1=6 \end{equation*} \]