Sachchidanand Prasad

13-02-2023

Problem: Prove tha the following series converges. \[ \sum_{n=1}^{\infty} \frac{n}{n^4 - n^{2} + 1}. \]

Solution: Note that for $n\ge 2$, we have \begin{align*} & n^4 - n^2 + 1 > n^4 - n^{2} = n^2 \left( n^2 - 1 \right) > n^2\times \frac{n^2}{2} > 0. \end{align*} Therefore, we have \begin{align*} & 0 \lt \frac{1}{n^4 - n^2 + 1} \lt \frac{2}{n^4} \\ \implies & 0 \lt \frac{n}{n^4 - n^{2} +1} \lt \frac{2}{n^3}. \end{align*} We will now use the comparison test for the convergence of the given series.

(Comparison Test) Suppose we have two series, $\sum a_n$ and $\sum b_n$ with $a_n,b_n \geq 0$ for all $n$ and $a_n \le b_n$ for all $n$ (we can also assume that there exists $n_0\in \mathbb{N} $ such that this holds for all $n\ge n_0$). Then,
  1. If $\sum_{n=1}^{\infty} b_n$ is convergent, then $\sum_{n=1}^{\infty} a_n$ is also convergent.
  2. If $\sum_{n=1}^{\infty} a_n$ is divergent, then $\sum_{n=1}^{\infty} b_n$ is also divergent.

Note that using the $p$ series test, which says that $\sum \frac{1}{n^p}$ converges if and only if $p>1$, the series \[ \sum_{n=1} ^ \infty \frac{2}{n^3} \] converges. Therefore, using the comparison test, the given series is convergent.