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$ \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots $

This is a pretty unorthodox problem, and I'm not quite sure how to simplify it. Could I get a solution? Thanks.

user84413
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Bob Joe
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6 Answers6

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$$\frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots=\sum_{n=1}^{\infty} \frac{2n-1}{2^n}$$

evinda
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    What does that mean? – Bob Joe Aug 19 '14 at 23:23
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    Rather, how do I calculate that? – Bob Joe Aug 19 '14 at 23:23
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    At the numerator we have all the odd numbers.The odd numbers are either of the form $2n-1$ or of the form $2n+1$.Since the first term of the sum is $1$,we take the form $2n-1$ (at $2n+1$ , $1$ is not included,since $n \in \mathbb{N}$) $$$$ At the denominator,we have all the powers of $2$ and these are of the form $2^n$. – evinda Aug 19 '14 at 23:29
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    Now you can split it up into $\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-\sum_{n=1}^{\infty}\frac{1}{2^n}$, and use $D_{x}\sum x^{n}=\sum nx^{n-1}$ in the first term. – user84413 Aug 19 '14 at 23:40
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    This is just providing alternative notation for the question, it does not answer it – M.M Aug 20 '14 at 11:09
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$$\sum_{n=1}^{+\infty}\frac{2n-1}{2^n}=-1+2\sum_{n=1}^{+\infty}\frac{n}{2^n}=-1+2\sum_{n=1}^{+\infty}\sum_{m\geq n}\frac{1}{2^m}=-1+4\sum_{n=1}^{+\infty}\frac{1}{2^n}=-1+4=3.$$

Jack D'Aurizio
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  • Can you explain a little bit more how you got from $1+2\sum_{n=1}^{+\infty}\sum_{m\geq n}\frac{1}{2^m}$ to $-1+4\sum_{n=1}^{+\infty}\frac{1}{2^n}$ – kleineg Aug 20 '14 at 12:48
  • The sum of a geometric progression with ratio $\frac{1}{2}$ is just twice the first term. – Jack D'Aurizio Aug 20 '14 at 19:27
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Another approach is to use summation by parts. This method works as follows. If $$S_N = \sum_{n=1}^{N} a_n b_n$$ then we can define $$B_n = \sum_{k=1}^{n} b_k$$ Then $$S_N = a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n)$$ In this problem, we can set $a_n = 2n-1$ and $b_n = 1/2^n$. Then $$B_n = \sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n}$$ and $a_{n+1} - a_n = 2$ for all $n$. Therefore, $$\begin{align}S_N &= a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n) \\ &= (2N-1)\left(1 - \frac{1}{2^N}\right) - \sum_{n=1}^{N-1} 2\left(1 - \frac{1}{2^n}\right) \\ &= 2N - 1 - \frac{2N - 1}{2^N} - 2(N-1) + 2\sum_{n=1}^{N-1} \frac{1}{2^n} \\ &= 1 - \frac{2N - 1}{2^N} + 2\sum_{n=1}^{N-1} \frac{1}{2^n} \\ \end{align}$$ The middle term goes to zero as $N \rightarrow \infty$, and the rightmost term goes to $2$. So in the limit, we get $1 + 2 = 3$.

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In this answer, I give a pictorial proof that $$\frac14+\frac28+\frac3{16}+\frac{4}{32}+\frac{5}{64}+\cdots=1$$ With your sum after dividing by $2$ we have $$\frac14+\frac38+\frac5{16}+\frac{7}{32}+\frac{9}{64}+\cdots$$ and the same picture can be used if we expand the original $\frac12\times\frac12$ square to the left as well as to the right and upward. The limit will fill up a rectangle that is $\frac32$ unit wide by $1$ unit tall. So we will have $\frac32$ square units of area. Since we had to halve your series to get this picture, your series sums to $3$.

Community
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2'5 9'2
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We can take this as the $x=\frac12$ case of the series

$$f(x)=x+3x^2+5x^3+7x^4+\cdots+(2n-1)x^n+\cdots = \sum_{n=1}^\infty (2n-1)x^n$$

We can recognize in this summation the geometric series

$$S(x)=x+x^2+x^3+\cdots= x+x\left(x+x^2+\cdots\right)=x+x S(x)=\frac{x}{1-x}$$ where in the last equality we have solved the rest of the equation for $S(x)$.

The obvious thing would be to pull this part of $f(x)$ out of the series above, but it'll actually save us some effort if we consider $f(x)-S(x)$:

$$f(x)-S(x)=2x^2+4x^3+6x^4+\cdots = 2x^2\left[1+2x+3x^2+\cdots\right]=2x^2\sum_{n=1}^\infty n x^{n-1}$$

Do the terms of the bracketed series remind you of anything from calculus?

Semiclassical
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$$\begin{align} \sum_{k=1}^\infty \frac {2k-1}{2^k} & =\frac {1}{2}+\frac {3}{2^2}+\frac {5}{2^3}+\frac {7}{2^4}+\frac {9}{2^5}+\cdots+\frac {2k-1}{2^k}+\cdots \\[1ex] 2\sum_{k=1}^\infty \frac {2k-1}{2^k} & = \sum_{k=1}^\infty \frac {2k-1}{2^{k-1}} \\ &= \sum_{j=0}^\infty \frac {2j+1}{2^j} \\ &= 1 + \sum_{j=1}^\infty \frac {2j+1}{2^j} \\[1ex] \sum_{k=1}^\infty \frac {2k-1}{2^k} & = 1 + \sum_{j=1}^\infty \frac {2j+1-(2j-1)}{2^j} \\ & = 1 + \sum_{j=1}^\infty \frac{1}{2^{j-1}} \\ & = \mathop{1 + \underbrace{\sum_{k=0}^\infty \frac{1}{2^{k}}}}_{\text{is this term familiar?}} \\[2ex] & = 1 + 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \cdots + \frac 1 {2^k} + \cdots \end{align}$$

Graham Kemp
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