# Constructing finite versions of arbitrary series

I've been wondering if/when it's possible to "truncate" a series.

Example 1

For example, the closed form for the series of naturals is:

$\frac{1}{(x-1)^2}$ = $1 + 2z + 3x^2 + \cdots$

The "truncated" form is:

$\frac{1-(n+1)x^n+n\cdot x^{n+1}}{(x-1)^2}$ = $1 + 2z + 3x^2 + \cdots + n\cdot x^{n-1}$

Example 2

The closed form of integers of the form $2^n$ is:

$\frac{1}{1-2x}$ = $1 + 2x + 4x^2 + \cdots + (2x)^n + \cdots$

The "truncated" form is:

$\frac{1 - (2x)^n}{1 - 2x}$ = $1 + 2x + 4x^2 + \cdots + (2x)^{n-1}$

The Search

Could anyone help with finding appropriate resources that address "truncating" a series? It would be great if there is some work that may include math that relates to finding truncated versions of series.

I don't have a very good background in math, so anything that doesn't require a lot of background would be very beneficial. However, I am very interested in learning, and ANY resource that touches upon this topic would be very beneficial and interesting for me.

I'd just like to know where I can learn more about this and related topics. Again, I'm very interested in being able to produce a closed form or recurrence for a "piece" of a series.

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Truncating to the first $n$ terms is the same as taking the Hadamard product with $\frac{1 - x^n}{1 - x}$, so if you can compute the corresponding integral then you're home free. In the special case that the original series is rational you can also do it using partial fraction decomposition, and then you are actually done by using appropriate generalizations of the two formulas you've already listed.
In general I see no reason for truncations to have nice closed forms. In particular I see no reason for the truncations of a series like $- \ln (1 - x)$ to have nice closed forms, since that would imply that the harmonic series has a nice closed form.
Mathematica gives the truncated portion as $x^n \Phi(x,1,n)$, with the special function being the Hurwitz–Lerch Transcendent. It is given $\Phi(z,s,\alpha)=\sum_{n=0}^\infty{\frac{z^n}{(n+\alpha)^s}}$. I thought I just ran across a way to rewrite a similar summation in terms of the Gamma function, but I'm probably way off somehow:-( If only these summations could be rewritten as a recursion... –  Matt Groff Nov 5 '10 at 4:30