The Arctangent Series From Termwise Integration

From a Geometric Series to an Arctangent Series

A geometric series can produce more than a formula for a rational function. Integrating its terms can produce a power series for an antiderivative. For the function \(1/(1+x^2)\), that antiderivative is \(\arctan x\), and the resulting series gives a useful representation of the inverse tangent near zero.

The integration has to be handled carefully. The geometric series is valid when \(|x|<1\), while the resulting arctangent series also converges at \(x=-1\) and \(x=1\). Those endpoint values require separate justification; convergence of the geometric series at the endpoints is not available.

The Geometric Series to Be Integrated

For a real number \(r\) with \(|r|<1\), the geometric-series formula is \(\frac{1}{1-r}=\sum_{n=0}^{\infty}r^n\). Set \(r=-x^2\). If \(|x|<1\), then \(|-x^2|<1\), so

The derivative of \(\arctan x\) is \(1/(1+x^2)\), and \(\arctan 0=0\). Integrating the displayed series from \(0\) to \(x\) therefore suggests integrating each term. Here is why that operation is justified for \(|x|<1\).

Fix any \(a\) with \(0<a<1\). For every \(t\in[-a,a]\), the absolute value of the tail after the \(N\)th term is bounded by

Since \(a<1\), this bound tends to zero as \(N\) increases, independently of \(t\). Thus the geometric series converges uniformly on \([-a,a]\). A uniformly convergent series of continuous functions can be integrated term by term on a closed bounded interval, so the suggested integration is valid whenever \(x\) lies inside such an interval. Every \(x\) with \(|x|<1\) lies in some interval \([-a,a]\) with \(|x|<a<1\).

Deriving the Maclaurin Series

Integrating from \(0\) to \(x\), term by term, gives the power series. Each term is integrated using the power rule, and the constant of integration is fixed by \(\arctan 0=0\).

The construction also provides a check on the formula. Inside \((-1,1)\), the series may be differentiated term by term on any smaller closed interval. Its derivative is

The series itself is zero at \(x=0\). Since its derivative agrees with the derivative of \(\arctan x\) throughout \((-1,1)\), and both functions have value zero at \(0\), they agree throughout that interval. This checks both the differentiated expression and the integration constant.

Convergence Inside the Interval

For a fixed \(x\) with \(|x|<1\), the absolute values of the terms are \(\frac{|x|^{2n+1}}{2n+1}\). They are bounded above by \(|x|^{2n+1}\), whose sum is a convergent geometric series. Therefore the arctangent series converges absolutely for \(|x|<1\). The derivation above establishes that its sum there is \(\arctan x\).

At the endpoints, the terms no longer have geometrically decreasing magnitudes. We must check those values separately, both to establish convergence and to identify the sums.

The Endpoint \(x=1\)

Substituting \(x=1\) formally gives the alternating series

Its term magnitudes \(1/(2n+1)\) decrease to zero, so the alternating-series test proves convergence. To prove that its sum is \(\arctan 1=\pi/4\), use the finite geometric identity before taking a limit. For every nonnegative integer \(N\) and \(0\leq t\leq1\),

Integrating from \(0\) to \(1\) gives the exact relation

The magnitude of the remaining integral is at most \(\int_0^1t^{2N+2}\,dt=1/(2N+3)\), because \(1+t^2\geq1\). This bound tends to zero as \(N\) increases. Therefore the partial sums tend to \(\pi/4\), which proves the endpoint identity.

The Endpoint \(x=-1\)

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

Its term magnitudes again decrease to zero, so the alternating-series test proves convergence. The series is also the negative of the series at \(x=1\). Its sum is therefore \(-\pi/4\), which equals \(\arctan(-1)\). This equality can also be checked directly from oddness: \(\arctan(-x)=-\arctan x\).

Interval of Convergence and a Common Pitfall

The series converges absolutely when \(|x|<1\), and it converges conditionally at both \(x=-1\) and \(x=1\). If \(|x|>1\), its terms do not approach zero: their magnitudes are \(|x|^{2n+1}/(2n+1)\), which grow without bound for sufficiently large \(n\). A convergent series must have terms tending to zero, so the series diverges there. Its interval of convergence is therefore \([-1,1]\).

A common mistake is to assume that the original geometric series can be substituted at \(x=1\) or \(x=-1\), and then integrated term by term there. At either endpoint, the geometric series for \(1/(1+x^2)\) does not converge: its terms have magnitude \(1\). The correct argument is to integrate the finite geometric identity and show that its integrated remainder tends to zero. That separate step establishes the endpoint values without applying the geometric-series formula outside its range of convergence.

The same care is needed with alternating-series error signs. A partial sum ending in a positive term lies above the limit; one ending in a negative term lies below it. Check the sign of the next term before writing a two-sided remainder bound.

Why the Construction Is Useful

The series gives a polynomial-based way to approximate inverse tangent values using only powers, fractions, and addition. Its alternating terms also provide a direct error bound: after a partial sum, the error has the sign of the first omitted term and magnitude less than that term, provided the term magnitudes decrease to zero. At \(x=1\), this supplies rational bounds for \(\pi/4\), and hence for \(\pi\).

The central reasoning has two parts. Uniform convergence on every closed interval strictly inside \((-1,1)\) justifies integrating the geometric series to obtain the function’s series. At the endpoints, an alternating-series argument proves convergence, while the integrated finite geometric identity identifies the sums. Keeping those arguments distinct avoids extending a formula beyond the conditions that justify it.