# Find the limit of $\lim_{n\rightarrow\infty}(\frac{1}{2}+\frac{3}{2^2}+…+\frac{2n-1}{2^n})$

Can anybody help me with this limit? $$\lim_{n\rightarrow\infty}\left(\frac{1}{2}+\frac{3}{2^2}+\cdots+\frac{2n-1}{2^n}\right)$$ Abel's transform wouldn't work on this one I guess, because another serie diverges As this contains a geometric series with ratio $1/2$, I had multiplied whole series by $1/2$ and then subtracted from original, but got original series back. Any ideas how can the limit be calculated?

• Almost sure it is a duplicate. Anyway, $$S_n-\frac{1}{2}S_n = \frac{1}{2}+\frac{2}{2^2}+\ldots+\frac{2}{2^n}-\frac{2n-1}{2^{n+1}}.$$ – Jack D'Aurizio Jan 9 '15 at 21:39
• See this for a (very) similar problem. – David Mitra Jan 9 '15 at 21:53

$S \ \ = \ \ \ \ \ \ \ \dfrac{1}{2}+\dfrac{3}{2^2}+\dfrac{5}{2^3}+\dfrac{7}{2^4}+\cdots$.

$2S = 1+\dfrac{3}{2}+\dfrac{5}{2^2}+\dfrac{7}{2^3} + \dfrac{9}{2^4} + \cdots$

Now, subtract the first equation from the second, but match up terms with the same denominator:

$S = 1+ \dfrac{2}{2} + \dfrac{2}{2^2} + \dfrac{2}{2^3} + \cdots$

This becomes $1$ plus a geometric series, which is easy to evaluate.

• thank you, seem's I chose a wrong pairs in subtraction, so I was stuck – shcolf Jan 9 '15 at 21:43
• Nit: Doesn't this assume the limit already exists? – Aryabhata Jan 9 '15 at 21:45
• ^Yes it does. Of course, convergence of this series is easy to prove using the standard tests learned in an introductory calculus course. – JimmyK4542 Jan 9 '15 at 21:54
• @JimmyK4542: Not just that, you are taking a term by term difference etc... (just pointing out for the readers, btw). – Aryabhata Jan 9 '15 at 23:05

We have: $$\lim_{n\to\infty}\left(\frac{1}{2}+\frac{3}{2^2}+\cdots+\frac{2n-1}{2^n}\right) = \sum_{n=1}^{\infty}\frac{2n-1}{2^n} = \sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-\sum_{n=1}^{\infty}\frac{1}{2^n}.$$ The sum on the right is a standard geometric series. The sum on the left can be determined by taking the usual geometric series: $$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n,\quad |x|<1$$ differentiating both sides and multiplying by $x$ after. I can expand further if you get stuck.

prove with induction that $$\sum_{i=1}^{n}\frac{2i-1}{2^i}=-2\, \left( 1/2 \right) ^{n+1}-4\, \left( 1/2 \right) ^{n+1} \left( n+ 1 \right) +3$$

• "prove with induction" No, and this advice is absurd, pedagogically speaking, as was already explained to this user. – Did Jan 11 '15 at 13:17
• but with the explicit sum he could solve the problem also. Prove with induction is very usefull in this case. What do you mean with pedagogically speaking? this is useless, since every teacher has another opinion. – Dr. Sonnhard Graubner Jan 11 '15 at 13:22
• We already played this game and you already evaded the problem the first time. "Prove with induction is very useful[l] in this case." Much less useful than solving the question as it is asked. The question does not ask for the partial sums but for the sum of the series. "every teacher has another opinion" Absolute relativism is classically invoked to mask that one seeks to avoid rational inquiry. Of course "with the explicit sum [they] could solve the problem also" but this is not the approach one should advocate first. – Did Jan 11 '15 at 13:30
• why not? in this case exists an explicit formula in the most of our cases it doesn't – Dr. Sonnhard Graubner Jan 12 '15 at 19:15
• Somehow I knew you would not listen. – Did Jan 12 '15 at 20:13

You can generalize it to: \begin{align*}\sum_{n=0}^\infty\frac {2n+1}{2^{n+1}}x^{2n}&=\left(\sum_{n=0}^\infty\frac 1{2^{n+1}}x^{2n+1}\right)'=\left(\frac x2\sum_{n=0}^\infty\frac{x^{2n}}{2^n}\right)'=\left(\frac x2\cdot\frac 1{1-x^2/2}\right)'=\left(\frac x{2-x^2}\right)'\\&=\frac{2-x^2+2x^2}{(2-x^2)^2}=\frac{2+x^2}{(2-x^2)^2}\end{align*} This holds for $|x^2/2|<1\iff |x|<\sqrt2$, so you can get the answer by setting $x=1$: $$\frac{2+1}{(2-1)^2}=3$$

In the same spirit as user2345215's answer, let us consider $$S=\sum_{i=1}^{\infty}(2i-1)x^i=2x\sum_{i=1}^{\infty}ix^{i-1}-\sum_{i=1}^{\infty}x^{i}=2x\frac{d}{dx}\Big(\sum_{i=1}^{\infty}x^{i}\Big)-\sum_{i=1}^{\infty}x^{i}$$ and $$\sum_{i=1}^{\infty}x^{i}=-1+\sum_{i=0}^{\infty}x^{i}=-1+\frac{1}{1-x}=\frac{x}{1-x}$$ So, compute the derivative, get the expression and replace $x$ by $\frac 12$.

• You might have missed CC0607's answer; it's even closer to yours. – epimorphic Jan 12 '15 at 4:11