Turning a Power Series Back into a Function

A power series is usually introduced as a way to build a function out of coefficients. The reverse reading is just as useful: a series that is written out term by term is often a familiar function in disguise, and recognising it converts an infinite sum into something you can evaluate, differentiate or integrate in one step.

The geometric series is the seed

Almost every identification starts from

$\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}\qquad\text{for }|x|<1.$

Two manipulations extend this single fact to a large family.

Substitution and factoring. Replacing $x$ by a suitable expression, or factoring a common power out of a series, puts a new series into geometric form. A series in $x^{2}$, for instance, is geometric with ratio $x^{2}$ and converges on $|x|<1$, not on the same interval as the original.

Term-by-term calculus. Inside the interval of convergence a power series can be differentiated and integrated term by term, and the radius of convergence is unchanged. Differentiating the geometric series gives

$\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^{2}}=\sum_{n=1}^{\infty}n x^{n-1}=\sum_{n=0}^{\infty}(n+1)x^{n},$

so a series with a linear coefficient is a perfect square reciprocal, not a logarithm. Integrating instead gives

$\int_{0}^{x}\frac{dt}{1-t}=-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n},$

which identifies a series with a reciprocal coefficient as a logarithm. A coefficient that contains $n!$ points toward the exponential or a trigonometric function, which is where the standard Taylor series come from.

Why index shifts matter

Two series that differ only by where the index starts describe the same function, and the shift is what makes a pattern visible. Writing $\sum_{n=1}^{\infty}n x^{n-1}$ as $\sum_{n=0}^{\infty}(n+1)x^{n}$ is the step that turns "coefficient $n$" into "coefficient $n+1$", which is what matches the derivative above. Re-indexing is bookkeeping, but skipping it is the most common reason a correct series fails to look like a known function.

A numeric spot check

Every identification can be verified at a single interior point, and doing so costs almost nothing. For $\sum_{n=1}^{\infty}\dfrac{x^{n}}{n}$ at $x=-0.5$: the closed form gives $-\ln(1.5)\approx-0.405$, and the partial sums $-0.5$, $-0.625$, $-0.666$, $-0.681$, $\dots$ move toward that value. If the closed form and the partial sums disagree, the signs or the starting index are wrong — a much faster diagnosis than re-deriving the whole series.

The interval is part of the answer

A power series and its closed form agree only where the series converges. The function $\dfrac{1}{1-x}$ is defined for every $x$ except $x=1$, but it equals the geometric series only on $|x|<1$, and outside that interval the equation is simply false. Endpoint behaviour is a separate question again: term-by-term differentiation and integration preserve the radius of convergence, but they do not preserve convergence at the endpoints, so each endpoint has to be checked on its own if the interval is closed.