Determine the convergence of the series (-1)n / (log (en + e-n))

Consider the series

$\sum_{n=1}^{\infty} \frac{(-1)^n}{\log(e^n + e^{-n})}.$

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The given series converges conditionally.

Proof. First, we have

$\lim_{n \to \infty} \frac{1}{\log(e^n + e^{-n})} < \lim_{n \to \infty} \frac{1}{\log (e^n)} = \lim_{n \to \infty} \frac{1}{n} = 0.$

Furthermore, the sequence of terms $a_n = \frac{1}{\log(e^n + e^{-n})}$ is monotonically decreasing. We can see this by looking at

$f(x) = \frac{1}{\log(e^x+ e^{-x})} \quad \implies \quad f'(x) = \frac{e^{-x} - e^x}{(e^x+e^{-x})(\log (e^x + e^{-x}))^2}.$

This derivative is negative for all $x > 0$ (since the numerator is negative and the denominator is always positive). Hence, the sequence $\{ a_n \}$ is monotonically decreasing. Therefore, by the Leibniz rule we have

$\sum_{n=1}^{\infty} \frac{(-1)^n}{\log(e^n + e^{-n})}$

converges.

This convergence is conditional since

$\sum_{n=1}^{\infty} \left| \frac{(-1)^n}{\log(e^n + e^{-n})} \right| = \sum_{n=1}^{\infty} \frac{1}{\log(e^n + e^{-n})}.$

This series diverges by the comparison test since

$\frac{1}{\log (e^n + e^{-n})} \geq \frac{1}{\log(2e^n)} = \frac{1}{n + \log 2} \geq \frac{1}{2n},$

and

$\sum_{n=1}^{\infty}\frac{1}{2n} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n}$

diverges $. \qquad \blacksquare$

Stumbling Robot

Determine the convergence of the series (-1)ⁿ / (log (eⁿ + e^-n))

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