I am trying to calculate the sum of the following series:

$\sum_{k=1}^\infty (\frac{1}{2k}-\frac{1}{2k+1})$.

I know from Wolfram Alpha that the sum should be $1-\log(2)$, but the series is not telescopic and I am not able to reduce the general term to something familiar with the Taylor expansion of the logarithm.

Do you have any hint on how to do it?

  • $\begingroup$ What is the Taylor expansion of the logarithm? For example $\log(1+x)=x-\frac12x^2+\frac13x^3-\frac14x^4+\cdots$ where it converges, and letting $x=1$ and looking at partial sums resembles partial sums of $\sum (\frac{1}{2k}-\frac{1}{2k+1})$ $\endgroup$
    – Henry
    Apr 15 at 12:37

1 Answer 1


Write out a few terms: $$\newcommand{\r}[2]{\color{red}{\frac{#1}{#2}}} \newcommand{\b}[2]{\color{blue}{\frac{#1}{#2}}} \sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{2k+1} = \b 1 2 - \r 1 3 + \b 1 4 - \r 1 5 + \b 1 6 - \r 1 7 + \cdots $$ With this in mind, we can clearly rewrite the series as $$ \sum_{n=2}^\infty \frac{(-1)^{n}}{n}= 1+\sum_{n=1}^\infty \frac{(-1)^{n}}{n}= 1 - \ln 2 $$

  • 1
    $\begingroup$ I was about to say, "But wait! This series only converges conditionally. Aren't you rearranging its terms and assuming the sum is the same?" But you're not rearranging terms, of course; you're writing the same terms in the same order with a different formula. $\endgroup$ Apr 15 at 15:23
  • $\begingroup$ lol yeah I fretted over that myself for a hot minute after posting. $\endgroup$ Apr 15 at 22:00

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