# Sequence Sum {1/2 + 1/4 + 1/6 +…} to infinite

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?

• 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

It doesn't. Up to a factor of $\frac12$ per summand this is the harmonic series, the standard example for divergence.
$$S = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \;...$$
$$[Parentheses \; have \; been \; put \; for \; comparing \; the \; series \; S \; and \; T]$$ 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.
• $\frac{1}{8}+\frac{1}{8}+\frac{1}{8} = \frac{1}{2}$? – Dilip Sarwate Sep 10 '12 at 11:43