Please could someone explain to me how to find out whether the following infinite series absolutely converges, diverges or conditionally converges:


Am I allowed to say that $s_n = \frac{1}{2n} - \frac{1}{2n-1}$ whose limit is 0, meaning the series converges to 0? But then $\sum\frac{1}{n}$ would be a counter example to this kind of logic so I am wrong I think. I haven't worked with series much before so I am not sure what things to rely on.


  • 1
    $\begingroup$ If you want to determine the sum you can rewritte it as $\sum_{n=1}^\infty \frac{1}{2n} - \frac{1}{2n-1} = \sum_{n=1}^\infty \frac{(-1)^n}{n} = - \log(2)$ where the last sum follows form the Taylor series for $\log(1+x)$ $\endgroup$ – Winther Jan 2 '17 at 20:38

Because $\frac{1}{n+(-1)^{n+1}}\geq \frac{1}{n+1}$ the series of absolute values is bounded from below by the harmonic series which is divergent.

The series is, however, conditionally convergent which can be seen adding each two adjacent element to form a new series. The value of it is easily found, to be $-\log (2)$.


After finishing my posting I noticed that Winther had already obtained the result - Log(2).

  • $\begingroup$ The series is not alternating (the signs are allright but the absolute values do not decrease). $\endgroup$ – Did Jan 2 '17 at 21:38
  • $\begingroup$ @Did: there's a slight confusion of terms: I was referring to the elements a(n) which are are alternating in sign and going to zero (condition for Leibniz criterion). The sum (series) a(1) + a(2) + ... is then non-monotonously going to - log(2). $\endgroup$ – Dr. Wolfgang Hintze Jan 2 '17 at 21:49
  • $\begingroup$ Precisely, this series does not satisfy Leibniz criterion: en.wikipedia.org/wiki/Alternating_series_test#Formulation $\endgroup$ – Did Jan 2 '17 at 22:08
  • $\begingroup$ You are right. The series is not convergent by the Leibniz criterion. I'll correct that in my post. $\endgroup$ – Dr. Wolfgang Hintze Jan 2 '17 at 22:49

The series conditionally converges, because the sum of two consecutive terms is $1/(n^2-1).$ It does not converge absolutely, by comparison with the sum of $1/(n+1).$

  • $\begingroup$ Seems to be a typo: the sum of two consecutive terms is $-\frac{1}{n^2 + n}$ or $-\frac{3}{n^2+n-2}$ for $n$ odd and even respectively. $\endgroup$ – Winther Jan 2 '17 at 21:01
  • $\begingroup$ @Witnther True, I was doing it in my head and typing too quickly. $\endgroup$ – Igor Rivin Jan 2 '17 at 21:06

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