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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?

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  • $\begingroup$ 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 . $\endgroup$ – Chinmaya mishra Aug 14 '18 at 16:09
  • $\begingroup$ Ok, is that common to do? $\endgroup$ – bubububub Aug 14 '18 at 16:17
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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.

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As In series, finite number of terms does not make any change in convergence. So even if for some term, a principle is not working then you can remove that term and do an analysis of remaining infinitely many terms.
As in your case, you can remove first term.
Remaining part as Santos Sir Provided.

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