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i have came across a series, i am trying to find its sum knowing the fact that, if it converges and its common ratio ex. r is: -1 < r < 1, then i can use the specified formula $\frac{a}{1-r}$ , which specifically means first term of series over 1 minus common ratio

here is the series $\sum_{n=1}^{\infty}\frac{2n-1}{2^n}$

i manipulated it this way to prove its convergence: $\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\frac{1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\left(\frac{1}{2}\right)^n$

$\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

using it i get the result 1, which actually should be 3

user122673
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  • See this for how to deal with a similar series. (Write your sum as $\sum\limits_{n=1}^\infty {2n\over2^n}-\sum\limits_{n=1}^\infty {1\over2^n}$.) – David Mitra Jan 20 '14 at 23:00

3 Answers3

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$$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}\frac{2n}{2^n}-\sum_{n=1}^{\infty}\frac{1}{2^n}=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-1.$$

We use Maclaurin series for function $\frac{1}{(1-x)}$ $$\frac{1}{1-x}=1+x+x^2+x^3+\dots = \sum_{n=0}^{\infty} x^{n}$$ Differentiating both sides of this equation we get that $$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\dots = \sum_{n=1}^{\infty}n x^{n-1}$$ if $x=\frac 12$ then $\sum_{n=1}^{\infty}n (\frac 12)^{n-1}=\frac{1}{(1-\frac 12)^2}=4$ $$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-1=3$$

nadia-liza
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  • I am also curious to know the explanation why it wont work with the general geometric sum formula ($\frac{a}{1-r}$) if its proven that the series is convergent, but it works that way! – user122673 Jan 21 '14 at 18:11
  • @user122673 Sequence ${\frac{1}{2^n}}$ is geometric progression. But sequences ${\frac{2n+1}{2^n}}$ and ${\frac{n}{2^{n-1}}}$ are not geometric progression (See http://en.wikipedia.org/wiki/Geometric_progression). Thus we can not use the specified formula $\frac{a}{1−r} to them. I am sorry, maybe I do not quite understand your question. (I did not speak English). – nadia-liza Jan 21 '14 at 20:58
  • Well what i was trying to ask is, if i would be able to use the method above for other similar series or for all similar series, would there be any case it would not work – user122673 Jan 21 '14 at 21:58
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Since you know it converges, let $S=\sum_{n=1}^\infty\frac{2n-1}{2^n}$. Then we have $$\begin{align} S &=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\cdots&\text{so}\\ \frac{1}{2}S&=\phantom{\frac{1}{2}}+\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\cdots&\text{subtracting, we get}\\ S-\frac{1}{2}S&=\frac{1}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\cdots&\text{so}\\ S-\frac{1}{2}S+\frac{1}{2}&=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\cdots&\text{so}\\ &=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots = 2 \end{align}$$ and thus we have $$ \frac{1}{2}S+\frac{1}{2}=2 $$ from which we can conclude that $S=3$.

This idiom, suitably manipulating a series as a linear combination leading to known sums, is quite often handy.

Rick Decker
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A helpful trick in such problems is the following: Introduce a variable $x$ to convert your "constant series" into a power series in $x$. This gives access to operations like differentiation or integration which then result in a series you can sum to a simple function you are familiar with. Looking at the given series I'd introduce the as yet unknown function $f$ defined by $$f'(x)=\sum_{n=1}^\infty {(2n-1)x^{2n-2}\over 2^n}\ .$$ Then, up to an additive constant, $$f(x)=\sum_{n=1}^\infty{x^{2n-1}\over 2^n}={1\over x}\sum_{n=1}^\infty\left({x^2\over 2}\right)^n={x^2/2\over x(1-x^2/2)}={x\over 2-x^2}\ .$$ It follows that $$f'(x)={1\over 2-x^2}-{x(-2x)\over(2-x^2)^2}\ ,$$ which implies $f'(1)=3$, and this is your sum.

Christian Blatter
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