# Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc.

I need to find limit of sequence $$\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)$$ I tried to solve it and stopped here $$f(n+1) = \frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}+\frac{2n+1}{2^{n+1}}$$ $$2f(n+1) = 1+\frac{3}{2}+\frac{5}{2^{2}}+\cdots+\frac{2n-1}{2^{n-1}}+\frac{2n+1}{2^{n}}$$ $$2f(n+1) -f(n) = 1+ \left(1 + \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\right) = 1 + g(n)$$ I can find the limit of $g$, but what to do with the other parts?

• Note that $\frac{2n}{2^{n}} = \frac{d}{dt} t^n |_{t=1/2}$ so you might be able to use power series since we have $\frac{2n-1}{2^n}=\frac{2n}{2^{n}} -\frac{1}{2^n}$. – Surb Apr 11 '18 at 11:36
• @Surb I need to calculate it without derivatives – amplifier Apr 11 '18 at 11:40
• Then you should clearly state it in your post. – Surb Apr 11 '18 at 11:41
• @Surb changed the post – amplifier Apr 11 '18 at 11:43
• – lab bhattacharjee Apr 11 '18 at 11:46

Let's assume that the limit exists. If we call the limit $s$ then

$s = \sum_1^{\infty}\frac{2n-1}{2^n}$

$\Rightarrow 2s = 1 + \sum_1^{\infty}\frac{2n+1}{2^n} = 1 + \sum_1^{\infty}\frac{2n-1}{2^n} + \sum_1^{\infty}\frac{2}{2^n}$

$\Rightarrow 2s = 1 + s + \sum_0^{\infty}\frac{1}{2^n}$

$\Rightarrow 2s = 1 + s + 2$

$\Rightarrow s=3$

So if the limit exists then it must be 3.

Now you just have to prove that the limit exists i.e. the series converges.

• To prove convergence: Using something similar to OP's notation $2f_{n+1} - f_n = 1 + g_n \le 3$. Thus by induction, if $f_n \le 3$ then $f_{n+1} \le 3$. $f_n$ is monotonically increasing and bounded above, so is convergent. – Aryabhata Apr 12 '18 at 1:43

Using the identity $$\frac{2k+1}{2^{k-1}}-\frac{2k+3}{2^k}=\frac{2k-1}{2^k}\tag1$$ the sum can be written as a telescoping series \begin{align} \sum_{k=1}^\infty\frac{2k-1}{2^k} &=\sum_{k=1}^\infty\left(\frac{2k+1}{2^{k-1}}-\frac{2k+3}{2^k}\right)\\ &=3\tag2 \end{align}

$$\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}}\right)=\sum _1^{\infty} \frac {2n-1}{2^n}=2\sum _1^{\infty} \frac {n}{2^n} -\sum _1^{\infty} \frac {1}{2^n}=2\sum _1^{\infty} \frac {n}{2^n} -1$$

Note that

\begin{align} \sum _1^{\infty} \frac {n}{2^n}&=\left(\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\cdots\right)\\&=\left(\frac12+\frac14+\frac18 +\cdots\right) +\left(\frac14+\frac18+\frac1{16}+\cdots\right)+\left(\frac18+\frac1{16}+\frac1{32}+\cdots\right)+\cdots\\&=1+\frac12+\frac14+\frac18+\cdots=2 \end{align}

Thus we have $$\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)= 3$$

Use summation by parts. Given a sequence $f(n)$ define $(\Delta f)(n)=f(n+1)-f(n)$ and note that $$\sum_{n=a}^b(\Delta f)(n)=f(b+1)-f(a).$$ Further given two sequences $f(n)$ and $g(n)$, note that $$(\Delta fg)(n)=f(n+1)\Delta g(n)+g(n)\Delta f(n)$$ whence $$\sum_{n=a}^bg(n)\Delta f(n)=[f(b+1)g(b+1)-g(a)f(a)]-\sum_{n=a}^bf(n+1)\Delta g(n)\tag{1}$$ Let $h(n)=-2/2^n$, and note that $\Delta h(n)=2^{-n}$. So by (1) $$\sum_{n=1}^k\frac{2n-1}{2^n}=[(2k+1)(-2^{-k})-1(-1)]+2\sum_{k=1}^n2^{-k}.\tag{2}$$ Let $k\to \infty$ in (2) and use the formula for a geometric series to deduce that $$\sum_{n=1}^\infty\frac{2n-1}{2^n}=1+2=3$$