Deriving the Arctangent Series From the Geometric Series

From a Geometric Expansion to an Arctangent Series

The geometric series is useful not only for representing rational functions, but also for deriving power series for other functions. A particularly direct example begins with the derivative of the inverse tangent function: differentiating \(\arctan x\) gives a reciprocal quadratic expression that can be expanded geometrically.

Integrating that expansion produces an alternating series involving odd powers of \(x\). The endpoint behavior needs care: the original geometric series fails at its endpoints, even though the integrated series converges there. We will establish the representation, identify exactly where it holds, and use partial sums to approximate values.

The Geometric Expansion

For a real number \(r\) with \(|r|<1\), the geometric series satisfies

Set \(r=-x^2\). This substitution meets the convergence condition precisely when \(|x|<1\), and gives

Since \(\frac{d}{dx}\arctan x=\frac{1}{1+x^2}\), integrating the terms suggests

The constant of integration is zero because \(\arctan 0=0\). The geometric series itself only gives the reciprocal expansion for \(|x|<1\), so we must still justify integration and decide whether the resulting series represents \(\arctan x\) at \(x=1\) or \(x=-1\).

Why Integrating Produces the Series

One justification uses termwise integration of a power series inside its interval of convergence. On every closed interval contained in \((-1,1)\), the geometric series converges uniformly, so it may be integrated term by term there. The fundamental theorem of calculus then identifies the integral with \(\arctan x-\arctan 0\).

A finite identity gives a direct argument that also handles the endpoints. For any nonnegative integer \(N\), the finite geometric formula gives

Integrate from \(0\) to \(x\). The integral of the left side is \(\arctan x\). Integrating each of the finitely many polynomial terms gives the partial sum

Thus the exact difference between the function and this partial sum is the integral of the remainder:

If \(|x|\leq 1\), then \(1+t^2\geq 1\). For \(x\geq 0\), the absolute value of the remainder is therefore at most

For \(x<0\), reversing the limits and using \(|t|\leq 1\) gives the same bound with \(|x|\) in place of \(x\). In either case, the bound tends to zero as \(N\) tends to infinity. Hence the partial sums converge to \(\arctan x\) at every point of the closed interval \([-1,1]\).

Convergence and the Interval of Validity

The power series converges absolutely for \(|x|<1\). Indeed, the absolute values of its terms are bounded above by \(|x|^{2k+1}\), and the geometric series with ratio \(x^2\) converges when \(|x|<1\).

At \(x=1\), the series becomes

Its term magnitudes decrease to zero, so the alternating-series test gives convergence. It is not absolutely convergent: the sum of the absolute values is \(\sum_{k=0}^{\infty}1/(2k+1)\), which diverges. At \(x=-1\), the terms have the same magnitudes and also alternate, so the series likewise converges conditionally. The finite-remainder argument proves that these endpoint sums equal \(\arctan 1=\pi/4\) and \(\arctan(-1)=-\pi/4\), respectively.

For \(|x|>1\), the terms do not tend to zero. In fact, their magnitudes are \(|x|^{2k+1}/(2k+1)\), which grow without bound as \(k\) increases. Since a convergent series must have terms tending to zero, the power series diverges outside \([-1,1]\). Thus \([-1,1]\) is the interval where this series converges to the arctangent function; its interior is the interval of absolute convergence, and its two endpoints are included by conditional convergence.

Worked Example: Approximate \(\arctan(1/2)\)

Worked Example: Use the Endpoint \(x=1\)

Worked Example: Approximate \(\arctan(1/3)\)

A Common Endpoint Mistake

A tempting argument is to substitute \(x=1\) into the geometric expansion of \(1/(1+x^2)\), write \(1-1+1-1+\cdots\), and integrate that series to obtain the arctangent series at the endpoint. This is invalid: the geometric series requires \(|x^2|<1\), and at \(x=1\) its terms do not tend to zero. The same issue occurs at \(x=-1\).

The correction is to keep the finite geometric identity and integrate its remainder before taking a limit. At either endpoint, the absolute value of the integrated remainder is at most \(1/(2N+3)\), which tends to zero. This establishes the endpoint value without treating a divergent geometric series as if it converged.

A second practical check is to keep the signs and denominators paired correctly: integrating \((-1)^k x^{2k}\) gives \((-1)^k x^{2k+1}/(2k+1)\). The odd denominator comes from the exponent after integration, not before it.

Using the Series Responsibly

For a value \(x\) strictly between \(-1\) and \(1\), the terms decrease in magnitude, and their signs alternate. A finite sum therefore gives an approximation whose error is no greater than the magnitude of the first omitted term. This provides a useful accuracy guarantee without needing to know the exact value of \(\arctan x\).

At the endpoints, the same alternating-series estimate applies, although convergence is slower because the term magnitudes are \(1/(2k+1)\). Outside the interval, the terms fail to approach zero, so increasing the number of terms does not produce a valid approximation from this power series. Always check the input against the interval of convergence before using a power series.