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What is $|a_n|$ of the power series $\sum_{n = 1}^{\infty}\frac{(-1)^n}{2n + 1} x^{2n + 1}$?

A power series has the form $\sum_{n = 1}^{\infty}a_nx^n$, so $|a_n| = \frac{1}{2n + 1}$ in this case. My professor wrote $|a_n|$ as a piece-wise function:

$|a_n| = \begin{cases} 0 & \text{ if n even } \\ \frac{1}{n}{} & \text{ if n odd } \end{cases} $

I don't really understand how he got those?

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The powers of $x$ in your series is the set $\{1, 3, 5, 7, 9, ...\}.$ You see, it contains no even numbers. That's why you can say it's $|a_n| = 0$ if $n$ is even and $\frac{1}{n}$ if $n$ odd.

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.......hint: what are the coefficients of the terms $a_{2n}x^{2n}$ ?

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  • $\begingroup$ $\frac{1}{4n + 1}$? I don't really follow... $\endgroup$ – Adrian Feb 18 '15 at 7:24
  • $\begingroup$ I am still unclear on why the coefficients of even terms are 0. So $a_1 = -\frac{1}{3}, a_2 = -\frac{1}{3} + \frac{1}{5}, a_3 = -\frac{1}{3} + \frac{1}{5} - \frac{1}{7}$ and so forth...I am still not really seeing it $\endgroup$ – Adrian Feb 18 '15 at 7:37

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