Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
Did you mean the sequence $\{1/2\,,\,/14\,,...\,1/2k\,,...\}\,$ or the series $$\frac{1}{2}+\frac{1}{4}+...=\sum_{n=1}^\infty\frac{1}{2n}\,\,?$$ –  DonAntonio Sep 10 '12 at 8:40
    
I had meant the 'series', sorry for the incorrect wording. –  student101 Sep 10 '12 at 8:48
    
This sequence is clearly divergent! –  Wreza Shafaghi Oct 5 '12 at 20:29
add comment

2 Answers 2

It doesn't. Up to a factor of $\frac12$ per summand this is the harmonic series, the standard example for divergence.

share|improve this answer
add comment

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.