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Compute the sum of the series ∑ (((-1)n) / (2n+1)) * (x/2)2n

Find all x \in \mathbb{R} such that the series

    \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n} \]

converges and compute the sum.


First, we have

    \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n \left( \frac{x^2}{4} \right)^n}{2n+1}. \]

Then, applying the ratio test we have

    \begin{align*}  \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{ \left( \frac{x^2}{4} \right)^{n+1}}{2n+3} \right) \left( \frac{2n+1}{\left( \frac{x^2}{4} \right)^n} \right) \\[9pt]  &= \left( \frac{x^2}{4} \right) \left( \frac{2n+1}{2n+3} \right) \\[9pt]  &= \frac{x^2}{4}. \end{align*}

Thus, the series converges for all x with x^2 < 4 which implies |x| < 2. Furthermore, if x = \pm 2, then the series converges as the alternating harmonic series. Then, we compute the sum for |x| < 2,

    \begin{align*}  \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n} &= \left( \frac{2}{x} \right) \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n+1} \\[9pt]  &= \left(\frac{2}{x} \right) \arctan \left( \frac{x}{2} \right).  \end{align*}

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