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# Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?$an = n/n^2+1$

Solution

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Let's compare $a_n$ and $a_{n+1}$ in order to find whether the sequence is monotonic or not:

\begin{align*} a_n - a_{n+1} &= \frac{n}{n^2+1} - \frac{n+1}{(n+1)^2+1} \\ &= \frac{n\cdot((n+1)^2+1)-(n+1)(n^2+1)}{(n^2+1)\cdot((n+1)^2+1)} \\ &= \frac{n\cdot(n^2+2n+1+1)-(n^3+n+n^2+1)}{(n^2+1)\cdot((n+1)^2+1)} \\ &= \frac{n^3+2n^2+2n-n^3-n^2-n-1}{(n^2+1)\cdot((n+1)^2+1)} \\ &= \frac{n^2+n-1}{(n^2+1)\cdot((n+1)^2+1)} \end{align*}

Since $(n^2+1)\cdot((n+1)^2+1)$ is always positive, we need to evaluate the sign of the quadratic expression $n^2-n-1$ in order to evaluate the sign of the expression. For this, we need to find the zeros:

$\begin{equation*} n_{1,2}=\frac{-1\pm\sqrt{1+4}}{2} = \frac{-1\pm \sqrt{5}}{2} \end{equation*}$

Therefore, $n^2+n-1>0$ for $n>\frac{-1+\sqrt{5}}{2}$, that is for every $n\in\mathbb{N}$. Therefore, for an arbitrary $n$, we have:

$\begin{equation*} a_n - a_{n+1} > 0 \implies a_n > a_{n+1} \end{equation*}$

so the sequence is decreasing. That's why it is bounded by its first term:

$\begin{equation*} a_1 = \frac{1}{1^2+1} = \frac{1}{2} \end{equation*}$

To examine whether the sequence is bounded below, notice that it has a limit:

\begin{align*} \lim_{n \to \infty}\frac{n}{n^2+1}&=\lim_{n \to \infty}\frac{\frac{n}{n^2}}{\frac{n^2}{n^2}+\frac{1}{n^2}}\\ &=\lim_{n \to \infty}\frac{\frac{1}{n}}{1+\frac{1}{n^2}}\\ &=\frac{\lim_{n \to \infty}\frac{1}{n}}{\lim_{n \to \infty}1+\lim_{n \to \infty}\frac{1}{n^2}}\\ &=\frac{0}{1+0}=0 \end{align*}

Therefore, the sequence is decreasing and is bounded by 0 and $\frac{1}{2}$.

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