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Compute the sum of the series ∑ (2n xn) / n

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

    \[ \sum_{n=1}^{\infty} \frac{2^n x^n}{n} \]

converges and compute the sum.


First, we apply the ratio test,

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

Thus, the series converges for |x| < \frac{1}{2}. For the boundary, if x = -\frac{1}{2} then this is the alternating harmonic series which converges. If x = \frac{1}{2} then it is the harmonic series, which diverges. Therefore, the series converges for all x \in \left[-\frac{1}{2}, \frac{1}{2} \right). Then, we compute,

    \begin{align*}  && P(x) &= \sum_{n=1}^{\infty} \frac{(2x)^n}{n} \\[9pt]  \implies && P'(x) &= \sum_{n=1}^{\infty} (2x)^{n-1} \\[9pt]  &&&= \sum_{n=0}^{\infty} (2x)^n \\[9pt]  &&&= \frac{2}{1-2x} \\[9pt]  \implies && P(x) &= \int_0^x \frac{2}{1-2t} \, dt \\[9pt]  &&&= -\log(1-2x). \end{align*}

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