# Alternating series test [Question of monotone decreasing sequence]

I have to show that

Let $n\in\mathbb{N}\cup\{0\}$ and let $(a_n)_{n}$ be defined as $a_n=(-1)^n\frac{n}{n^2+1}$. Check convergence of the series $\sum_{n=0}^{\infty} a_n$.

Using the alternating series test, the series is convergent if $|a_n|$ is a monotone decreasing sequence and $\lim_{n\to\infty}|a_n|=0$.

Using induction to prove that $|a_n|$ is a monotone decreasing sequence, $P(1)$ would have me prove that $|a_0|\ge|a_1|$ since $n\in\mathbb{N}\cup\{0\}$, which is obviously not true: $$|a_0|=1\cdot 0=0\not\ge |a_1|=|(-1)\cdot \frac{1}{2}|=\frac{1}{2}$$ Though in the next question it is assumed that you have found that the series is convergent with the alternating series test, which would be true if $n\in\mathbb{N}$.

Is my induction hypothesis wrong?

• First term of the series is zero . If you include it , obviously the P(1) will fail since other terms are non-zero . So you apply induction to rest non-zero terms . – Chinmaya mishra Aug 14 '18 at 16:09
• Ok, is that common to do? – bubububub Aug 14 '18 at 16:17

It turns out that $\left(\frac n{n^2+1}\right)_{n\in\mathbb N}$ is decreasing. Therefore, the series $\displaystyle\sum_{n=1}^\infty(-1)^n\frac n{n^2+1}$ and so the series $\displaystyle\sum_{n=0}^\infty(-1)^n\frac n{n^2+1}$ converges too.