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Convergence of a series

Examine the convergence of a series

    \[\sum_{n=1}^{\infty} \left ( 1 - 2\exp \left ( \sum_{k=1}^{n} \frac{(-1)^k}{k} \right ) \right )\]

Solution

From Taylor’s theorem with integral remainder we have that

    \[-\ln 2 = \sum_{k=1}^{n} \frac{(-1)^k}{k} + (-1)^{n+1} \int_0^1 \frac{x^n}{1+x} \, \mathrm{d}x\]

However it is known that \displaystyle \int_{0}^{1} \frac{x^n}{1+x} \, \mathrm{d}x = \frac{1}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right ). Hence,

    \[\sum_{k=1}^{n} \frac{(-1)^k}{k} = -\ln 2 + \frac{(-1)^n}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

Exponentiating we get

    \[1- 2 \exp \left ( \sum_{k=1}^{n} \frac{(-1)^k}{k} \right ) = \frac{(-1)^n}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

Thus the series converges.

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