I was doing some exercises on the nature of infinite series when I came across this one intriguing series - $$\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{\ln(i+1)}.$$ I tried to solve it with D'Alembert's ratio test and came at the solution that this series is convergent as $\lim\limits_{n \to \infty}|\frac{u_{n+1}}{u_n}|$ comes out to be $\lim\limits_{n \to \infty}|\frac{-\ln(n+1)}{\ln(n+2)}| = \lim\limits_{n \to \infty}\frac{\ln(n+1)}{\ln(n+2)}$ which is smaller than 1 as $\ln x$ is strictly increasing. So, the $-$ sign doesn't really make any difference. But the answer in the textbook is that this series is conditionally convergent, meaning that it wouldn't have been convergent if not for the $-$ sign. Can anyone explain this to me?

  • $\begingroup$ What do you mean by $\sum_{i=1}^{\infty}\frac{(-1^{n-1})}{ln\enspace (i+1)}$ ? $\endgroup$ – S.Sundara Narasimhan Sep 14 '19 at 7:56
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    $\begingroup$ The limit is not smaller than $1$. It's smaller or equal to $1$. $\endgroup$ – Gae. S. Sep 14 '19 at 7:56
  • $\begingroup$ @Gae.S. so is it convergent? $\endgroup$ – Pratham Yadav Sep 14 '19 at 7:57
  • $\begingroup$ @Sil yes sorry, I'll edit that. $\endgroup$ – Pratham Yadav Sep 14 '19 at 7:58
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    $\begingroup$ Actually, by D'Alembert's test, the ratio limit converges to 1 (just try applying L'Hôpital's). To see that it is not absolutely convergent, consider the series $\sum_{i = 1}^\infty \frac{1}{\ln{i+1}}$. Now, as $\ln(x)$ grows in a smaller rate than $x$, we have that for some big natural number $n_0$, if $i > n_0$, then $\ln{i+1} < i + 1$, which implies that $\frac{1}{\ln{i+1}}> \frac{1}{i}$ . Comparison to the harmonic series gives us the conclusion that $\sum_{i = 1}^\infty \frac{1}{\ln{i+1}} $ is divergent - which implies that, if convergent, the series in your question does it conditionally $\endgroup$ – Nuntractatuses Amável Sep 14 '19 at 8:00

The series $\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{\ln(i+1)}$ is not absolutely convergent, because $0<\ln(i+1)<i+1$, hence $$\frac1{\ln(i+1)} >\frac 1{i+1},$$ and the harmonic series diverges.

However it is conditionally convergent by Leibniz' test since $\;\dfrac1{\ln(i+1)}\;$ is monotonically decreasing to $0$.


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