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What is the idea behind a closed form expression and what is the general way of finding the closed form solution of an infinite summation?

context: closed form solution of $\sum^\infty_{i=1}ia^i$.

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  • $\begingroup$ It really depends on the form of the expression. What close form do you know? $\endgroup$ – Jean-Sébastien Nov 11 '12 at 18:56
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Lets see:

$\begin{align} \sum_1^{\infty} ia^i &= a + a^2 + a^2 + a^3 + a^3 + a^3 + ......\\ &= \sum_1^{\infty} a^i + (\sum_2^{\infty} a^i) + (\sum_3^{\infty} a^i) + ..+ (\sum_{k+1}^{\infty} a^i)+.... \\ &= \sum_1^{\infty} a^i + a (\sum_1^{\infty} a^i) + a^2 (\sum_1^{\infty} a^i) + ..+ a^k (\sum_1^{\infty} a^i)+.... \\ &= \sum_1^{\infty} a^i (1+ a+a^2+....)\\ &= \sum_1^{\infty} a^i (1+\sum_1^{\infty} a^i) \\ &= \frac{a}{1-a}(\frac{1}{1-a}) \; \text{if a <1} \end{align}$

The trick is to look for patterns that repeat themselves.

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In general, finding the closed-form of a series or a finite summation is a difficult problem without a general way of attack. For your particular series, if I am correct in assuming that your $a$ is a constant, then let's begin with a geometric series $$\frac{1}{1-x} = \sum_{i=0}^\infty x^i$$ Let us now differentiate this series term by term to get $$\frac{d}{dx}\left(\frac{1}{1-x}\right)= \frac{1}{(1-x)^2}=\sum_{i=1}^\infty ix^{i-1}$$ Multiplying by $x$ now gives $$\frac{x}{(1-x)^2} = \sum_{i=1}^\infty ix^i$$ The above series will converge for $|x| < 1$, so simply substitute your desired value for $a$ to get the value of the series.

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Just for fun, I’ll note that a closed form for the summation $\sum_{k\ge1}kx^k$ can also be found without differentiation:

$$\begin{align*} \sum_{k\ge 1}kx^k&=\sum_{k\ge 1}\sum_{i=1}^kx^k\\\\ &=\sum_{i\ge 1}\sum_{k\ge i}x^k&&\text{reversing the order of summation}\\ &=\sum_{i\ge 1}\frac{x^i}{1-x}&&\text{sum of a geometric series}\\ &=\frac1{1-x}\sum_{i\ge 1}x^i\\ &=\frac1{1-x}\cdot\frac{x}{1-x}\\\\ &=\frac{x}{(1-x)^2}\;. \end{align*}$$

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The closed form in general will be difficult to obtain and how to obtain it will vary from problems to problems. In your case, for $a\in (-1,1)$, the geometric series is given by $$ \sum_{i=0}^\infty a^i=\frac{1}{1-a}, $$ series being $1$ if $a=0$.

Knowing this, you can see that $ia^{i-1}$ is the derivative of that series, so $\sum_{i=1}^\infty ia^{i-1}$ should be the derivative of $\frac{1}{1-a}$ with respect to $a$. From this, we get $$ \sum_{i=0}^{\infty}ia^i=a\sum_{i=0}^\infty ia^{i-1}=a\left(\frac{1}{1-a}\right)^{'}=\frac{a}{(1-a)^2} $$ Since our geometric series was valid for $a\in (-1,1)$, this one is also valid for that interval.

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  • $\begingroup$ What is wrong with $a=0$? $0^0$? $\endgroup$ – copper.hat Nov 11 '12 at 19:04
  • $\begingroup$ meh, good question, I guess i only wanted to avoid $0^0$ issues $\endgroup$ – Jean-Sébastien Nov 11 '12 at 19:05
  • $\begingroup$ Convention is your friend :-). $\endgroup$ – copper.hat Nov 11 '12 at 19:05
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I don't think there is a general procedure. However, analytic functions and their power series representations provide many useful equivalences.

Consider $f(x) = \frac{1}{1-x} = \sum_{k=0}^\infty x^k$, with $|x|<1$. This is analytic, so you can interchange differentiation and summation to get $f'(x) = \frac{1}{(1-x)^2} = \sum_{k=1}^\infty kx^{k-1}$ (again for $|x|<1$), and finally $xf(x) = \frac{x}{(1-x)^2} = \sum_{k=1}^\infty kx^{k-1}$.

However. I know this because someone showed to me (hundreds of years ago, or so). In general, some guesswork is required.

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