Reciprocal Powers From Geometric-Series Differentiation

From the Geometric Series to Reciprocal Powers

The geometric series is a basic power series with many uses in calculus. A particularly useful feature is that its sum, \(1/(1-x)\), can be differentiated. Repeating that operation produces series for higher reciprocal powers such as \(1/(1-x)^2\) and \(1/(1-x)^3\).

The coefficients change with each differentiation, but they follow a regular pattern. The key is to differentiate only where the original power series represents its function, and then reindex the result carefully.

The First Differentiation

For \(|x|<1\), the geometric series states that \[ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n. \] A power series can be differentiated term by term at every point strictly inside its interval of convergence. The geometric series has radius of convergence \(1\), so on \(|x|<1\) its derivative is the derivative of its sum.

The derivative of \(x^n\) is \(nx^{n-1}\) for \(n\ge1\). The constant term has derivative zero, so it is convenient to start the differentiated sum at \(n=1\):

Let \(k=n-1\), so \(n=k+1\). Then the same identity can be written with powers beginning at \(x^0\):

Repeated Differentiation and the General Formula

Continue differentiating. After a second differentiation of the geometric series, the left side is \(2(1-x)^{-3}\). On the right, each term \(nx^{n-1}\) differentiates to \(n(n-1)x^{n-2}\). Dividing by \(2\) and reindexing gives the series for \((1-x)^{-3}\).

The same pattern extends to every positive integer power. The \(r\)-th derivative of \((1-x)^{-1}\) is \(r!(1-x)^{-(r+1)}\). On the series side, differentiating \(x^n\) \(r\) times gives \(n(n-1)\cdots(n-r+1)x^{n-r}\) when \(n\ge r\). Reindex with \(k=n-r\), and divide by \(r!\).

Why the Coefficient Formula Works

Here is the full derivation. Set \(r=m-1\). Repeatedly differentiating the geometric identity \(r\) times is valid for \(|x|<1\), by the termwise differentiation theorem for power series. Differentiating the function gives \(r!(1-x)^{-(r+1)}\). Differentiating each series term gives the product of \(r\) descending factors shown below.

Replace \(n\) with \(k+r\). The exponent becomes \(k\), and the descending product becomes \((k+r)!/k!\). Dividing both sides by \(r!\) therefore gives

Finally, \(r+1=m\), so this is the stated formula. The reindexing matters: it turns the differentiated powers \(x^{n-r}\) into powers \(x^k\) beginning at zero, while the factorial quotient becomes exactly the binomial coefficient.

Convergence: The Endpoints Need Their Own Test

The formula obtained by differentiation is valid for \(|x|<1\). To determine the full interval of convergence, test \(x=1\) and \(x=-1\) separately. For a fixed positive integer \(m\), let \(c_k=\binom{k+m-1}{m-1}\). These coefficients are positive and satisfy \(c_k\ge1\).

At \(x=1\), the terms of the series are \(c_k\), so they do not approach zero. At \(x=-1\), the terms are \((-1)^kc_k\), whose absolute values are \(c_k\ge1\); these terms also do not approach zero. Since a convergent series must have terms tending to zero, the series diverges at both endpoints.

A Common Indexing Mistake

A frequent error is to differentiate the geometric series but forget that its powers and coefficients both change. For example, writing \(\sum_{n=0}^{\infty}nx^{n-1}\) after differentiation is not a sound way to display the result: its \(n=0\) term is undefined, and the derivative of the constant term should instead be zero.

The correction is to start the differentiated sum at the first index whose term is not constant, then reindex so the powers begin at zero. Another important restriction is that the termwise differentiation argument establishes the identity inside \(|x|<1\); endpoint convergence must be checked separately.

Why These Expansions Matter

Reciprocal-power series provide explicit polynomial approximations to functions that are not themselves polynomials. Their coefficients are easy to generate from the geometric series, and the exact convergence interval tells us where the infinite expansion represents the function. The same coefficient formula also makes it possible to read off any desired finite number of terms without repeatedly differentiating from the beginning.

The central method is compact: differentiate within the original interval of convergence, reindex with care, and test the endpoints rather than assuming they converge. For positive integer \(m\), the resulting coefficients are \(\binom{k+m-1}{m-1}\), and the series represents \((1-x)^{-m}\) precisely on \((-1,1)\).