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I've been told, the following series converges: $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2k}+\ldots$$

I can't get my head around, how to prove this converges; any hints?

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    $\begingroup$ Did you mean the sequence $\{1/2\,,\,/14\,,...\,1/2k\,,...\}\,$ or the series $$\frac{1}{2}+\frac{1}{4}+...=\sum_{n=1}^\infty\frac{1}{2n}\,\,?$$ $\endgroup$ – DonAntonio Sep 10 '12 at 8:40
  • $\begingroup$ I had meant the 'series', sorry for the incorrect wording. $\endgroup$ – student101 Sep 10 '12 at 8:48
  • $\begingroup$ This sequence is clearly divergent! $\endgroup$ – Wreza Shafaghi Oct 5 '12 at 20:29
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It doesn't. Up to a factor of $\frac12$ per summand this is the harmonic series, the standard example for divergence.

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Let

$$S = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \;...$$

$$[Parentheses \; have \; been \; put \; for \; comparing \; the \; series \; S \; and \; T]$$ Then the given series is simply $2S$. Hence in order to investigate the convergence of the given series, we have to investigate $S$ which is a well known series called the harmonic series and is a nice example of a series whose $t_n \rightarrow 0$ as $n \rightarrow\infty$ but the series is still divergent. To show it, we will construct another series $T$ such that $T < S$ and $T$ is divergent. Let $$T = 1+ \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4}\right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \; ...$$ It's easy to see that $T < S$. Also, the series $T$ is the same as the series $1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}+\; ... \; = \infty$. Hence $S$ is a divergent series as well and as a result, the given series in the question diverges.

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  • $\begingroup$ $\frac{1}{8}+\frac{1}{8}+\frac{1}{8} = \frac{1}{2}$? $\endgroup$ – Dilip Sarwate Sep 10 '12 at 11:43
  • $\begingroup$ I think you want four eighths (and correspondingly four terms in the last bracket on the first line). $\endgroup$ – tttppp Sep 10 '12 at 11:43
  • $\begingroup$ sorry for the typo. Have corrected it. $\endgroup$ – ajay Sep 10 '12 at 11:57

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