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  1. Given a function series, what are the ways often used to find the function that the series converges to?

    For example, how do you construct the function on the right hand side from the series on the left hand side: $$\sum_{n \in \mathbb{N} \cup \{ 0\}} (\begin{array}{c} 2n \\ n \end{array}) x^{n}=(1-4x)^{-1/2}$$ I suppose one may try to go the other way around, i.e. verify that the Taylor expansion of the function on RHS around 0 is the series on the LHS, but I was wondering if this is one of the acceptable way to construct the RHS from the LHS?

  2. Related questions:

    Given a sequence of real numbers, what are the ways to find the number the sequence converges to?

    What are the ways to find the number a given series converges to? For example,

    $$\sum_{n=0}^\infty \frac{1}{n!}=e$$

    and the series that converges to $\pi$.

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I don't think there is a standard approach. There are standard approaches to look for convergence, but not for what it is that is being converged to. It often takes some original thought. Famous example being Euler's discovery of $\sum_{n=0}^{\infty}\frac{1}{n^2}$. –  Raskolnikov Feb 18 '11 at 15:58
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The exercices we give our calculus students for homework lead them to think that most problems have a solution in terms of a simple formula or algorithm. While this is true in the case of multiplying polynomials or homogeneous differential equations with constant coefficients it is e.g. not true for "simple" sums that one might think of, like $\sum_{k=1}^\infty{1\over k^2}$. It is a miracle that this sum is not a "new constant" which only can be computed to so many places, but is something that can be expressed in terms of $\pi$, familiar to us from completely different circumstances. –  Christian Blatter Feb 18 '11 at 21:00
    
@Christian: how is the series related to $\pi$? How do you know that? –  Tim Feb 18 '11 at 21:04
    
This series was already mentioned in Raskolnikov's comment. The sum is $\pi^2/6$, a famous result of Euler. I don't know of a proof that could be given in a first calculus course. –  Christian Blatter Feb 19 '11 at 8:57

2 Answers 2

up vote 3 down vote accepted

In general most power series and most sequences do not have closed forms. The specific series you mention is hypergeometric and there are known algorithms for working with these, e.g. guessing or proving identities or algebraic or differential equations. I don't know much about this subject, but Petkovsek, Wilf, and Zeilberger might be a good place to start. (If you just want to get from the RHS to the LHS, use the generalized binomial theorem.)

I do not understand your second example. $\sum \frac{1}{n!} = e$ is more or less a definition. If you define $e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n$, then you can prove this by proving that $e^x = \sum \frac{x^n}{n!} = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$, which is a nice exercise.

There are special techniques which work on series that have a particular form, but again, there is no reason to expect that an arbitrary sum has a reasonable expression in terms of known constants.

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It depends on the function. In your first question you have the generating function for central binomial coefficients and a variety of techniques such as those described in generatingfunctionology. Your equation appears as (2.5.11)

In your second question, it is unclear exactly what you are asking, as the first equation is for me the definition of $e$.

Note that $\sum_{n=0}^\infty \frac{x^n}{n!}$ is its own derivative and so a multiple of $\exp (x)$, and looking at $x=0$ it must be $\exp (x)$, so back to your question with $x=1$ the value is $\exp(1)$ or $e$.

Or perhaps you want an approximation to the decimal value: for example the terms up to $n=9$ add up to 2.7182815... and it is easy to show that the others add up to less than 0.000001 so the sum is about 2.71828...

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