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Odd-Power Sums From Arctangent Representations

Use the arctangent series to identify odd-power sums and bound the error in their finite approximations.

Advanced 9 min read

What You'll Learn

  • Recognize alternating sums whose terms have odd powers and odd denominators.
  • Substitute a value into the arctangent series only when it lies in its convergence range.
  • Derive an explicit error bound after a finite number of terms.
  • Use lower and upper bounds to certify a rounded numerical answer.
  • Avoid common indexing and exponent errors when forming the series terms.

From an Odd-Power Sum to an Inverse Tangent

An alternating sum with odd powers in the numerator and odd integers in the denominator may look difficult to evaluate directly. The arctangent power series provides a systematic way to recognize such sums: if the powers and denominators match, the infinite sum is the value of an inverse tangent.

The substitution also gives more than an exact representation. The alternating terms decrease in magnitude, so the next omitted term bounds the error after truncation. This lets us determine whether a numerical approximation is accurate enough to round to a specified number of decimal places.

Key formula: For \(0\leq x\leq 1\), \[ \arctan x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{2k+1} =x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots. \] For \(0\leq x<1\), the series is absolutely convergent; at \(x=1\), it converges conditionally.

Why the Arctangent Series Works

A short derivation makes clear both the source of the terms and a useful error estimate. The finite geometric-series identity gives, for \(0\leq t\leq 1\),

$$ \frac{1}{1+t^2} = \sum_{k=0}^{m-1}(-1)^k t^{2k} + (-1)^m\frac{t^{2m}}{1+t^2}. $$

Integrate both sides from \(0\) to \(x\), where \(0\leq x\leq 1\). Since \(\arctan x=\int_0^x \frac{1}{1+t^2}\,dt\), this gives

$$ \arctan x = \sum_{k=0}^{m-1}(-1)^k\frac{x^{2k+1}}{2k+1} + (-1)^m\int_0^x\frac{t^{2m}}{1+t^2}\,dt. $$

The integral is the remainder after \(m\) terms. Its integrand is nonnegative, and \(1+t^2\geq 1\). Therefore,

$$ \left|\arctan x-\sum_{k=0}^{m-1}(-1)^k\frac{x^{2k+1}}{2k+1}\right| = \int_0^x\frac{t^{2m}}{1+t^2}\,dt \leq \int_0^x t^{2m}\,dt = \frac{x^{2m+1}}{2m+1}. $$

This proves the series formula: the bound tends to zero as \(m\) increases, including when \(x=1\). It also proves the remainder estimate directly. In the series, the terms alternate in sign and their magnitudes decrease to zero, so the usual alternating-series estimate gives the same bound: the error is at most the magnitude of the first omitted term.

Remainder bound: Define the \(m\)-term partial sum by \(S_m(x)=\sum_{k=0}^{m-1}(-1)^k\frac{x^{2k+1}}{2k+1}\). For \(0\leq x\leq 1\), \[ \left|\arctan x-S_m(x)\right|\leq \frac{x^{2m+1}}{2m+1}. \] For \(x>0\), the exact sum lies between \(S_m(x)\) and \(S_m(x)+(-1)^m\frac{x^{2m+1}}{2m+1}\).

Matching a Given Sum to the Formula

To use the series, compare the general term with the sum you want to evaluate. The denominator \(2k+1\) must match the odd powers \(x^{2k+1}\) for one fixed value of \(x\). For example, a numerator \(1/3^{2k+1}\) corresponds to \(x=1/3\); a numerator \(2^{2k+1}/5^{2k+1}\) corresponds to \(x=2/5\). Check the sign pattern and the starting index as well.

If \(0\leq x\leq 1\), substitution is justified by the series formula above. The endpoint \(x=1\) is included, though its series is only conditionally convergent. Values with \(x>1\) are outside this series’ convergence range, so this substitution does not evaluate a sum at those values.

Worked Example: Evaluate a Sum with Powers of One-Third

Consider \[ \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)3^{2k+1}}. \] Rewrite its general term as \[ \frac{(-1)^k}{(2k+1)3^{2k+1}} = (-1)^k\frac{(1/3)^{2k+1}}{2k+1}. \] Here \(x=1/3\), which lies between zero and one. The arctangent series therefore gives \[ \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)3^{2k+1}} =\arctan\left(\frac13\right). \]

To see the size of the terms, the first three give \[ S_3=\frac13-\frac{1}{3\cdot 3^3}+\frac{1}{5\cdot 3^5} =\frac13-\frac{1}{81}+\frac{1}{1215} =\frac{405-15+1}{1215} =\frac{391}{1215} \approx 0.3218106996. \] The next term has magnitude \[ \frac{(1/3)^7}{7}=\frac{1}{7\cdot 2187}=\frac{1}{15309}\approx 0.0000653210530. \] It is negative, so the sum lies between \(S_3-1/15309\) and \(S_3\): \[ 0.3217453785\ \lesssim\ \arctan\left(\frac13\right)\ \lesssim\ 0.3218106996. \] Both endpoints round to \(0.322\) to three decimal places, so this bound certifies that rounding. It does not certify four decimal places, because the endpoints round differently at that precision.

Check: The general formula identifies the sum as \(\arctan(1/3)\), and the next term is negative, consistent with the lower bound being \(S_3-1/15309\). The reciprocal and the displayed lower endpoint use \(1/15309\approx 0.0000653210530\).

Certifying a Numerical Approximation

An approximation is certified to a given number of decimal places when the entire error interval rounds to the same value at that precision. A small-looking next term is not, by itself, enough to claim a particular number of correct digits: compare both endpoints of the interval with the relevant rounding boundaries.

The remainder bound is especially convenient when \(x<1\), because successive terms decrease rapidly. The bound after \(m\) terms is the next term’s magnitude, \(x^{2m+1}/(2m+1)\). Adding or subtracting that amount from the partial sum gives a guaranteed interval containing the infinite sum.

Worked Example: Certify the Rounding of a Sum with \(x=2/5\)

Consider the odd-power sum \[ \sum_{k=0}^{\infty}(-1)^k\frac{2^{2k+1}}{(2k+1)5^{2k+1}}. \] Its general term is \((-1)^k(2/5)^{2k+1}/(2k+1)\), so the sum is \(\arctan(2/5)\). Use five terms: \[ S_5=\frac25-\frac{(2/5)^3}{3}+\frac{(2/5)^5}{5}-\frac{(2/5)^7}{7}+\frac{(2/5)^9}{9}. \] The first three terms combine as \[ \frac25-\frac{8}{375}+\frac{32}{15625} =\frac{17846}{46875}. \] The fourth term is \(128/546875\), so \[ S_4=\frac{17846}{46875}-\frac{128}{546875} =\frac{624226}{1640625}. \] The fifth term is \(512/17578125\), giving \[ S_5=\frac{624226}{1640625}+\frac{512}{17578125} =\frac{46820534}{123046875} \approx 0.3805097366. \]

The first omitted term is negative, with magnitude \[ \frac{(2/5)^{11}}{11} =\frac{2048}{537109375} \approx 0.0000038130. \] Consequently, \[ 0.3805059236\ \lesssim\ \arctan\left(\frac25\right)\ \lesssim\ 0.3805097366. \] Both endpoints round to \(0.381\) to three decimal places. Thus the five-term partial sum certifies the three-decimal approximation.

Check: The terms alternate, and the omitted term after \(S_5\) has index \(k=5\), so its sign is negative. The sum must therefore be below \(S_5\), as the interval shows.

Substitutions with Other Odd Powers

The value substituted need not be a unit fraction. If the powers appear as \(a^{2k+1}/b^{2k+1}\), then the appropriate value is \(x=a/b\), provided \(0\leq a/b\leq 1\). The same sum and remainder formulas apply. In particular, a finite partial sum gives a lower or upper bound depending on whether the next omitted term is negative or positive.

Worked Example: Recognize a Sum with Powers of Three-Fourths

Consider \[ \sum_{k=0}^{\infty}(-1)^k\frac{3^{2k+1}}{(2k+1)4^{2k+1}}. \] Since \(3^{2k+1}/4^{2k+1}=(3/4)^{2k+1}\), this is the arctangent series at \(x=3/4\), and hence the sum equals \(\arctan(3/4)\).

The first three terms are \[ \frac34-\frac{(3/4)^3}{3}+\frac{(3/4)^5}{5} = \frac34-\frac{9}{64}+\frac{243}{5120} = \frac{3363}{5120}. \] The next term is negative and has magnitude \[ \frac{(3/4)^7}{7}=\frac{2187}{114688}. \] Thus the exact sum lies in the interval \[ \frac{3363}{5120}-\frac{2187}{114688} \ \leq\ \arctan\left(\frac34\right)\ \leq\ \frac{3363}{5120}. \] The inequalities have the correct order because the first omitted term is negative.

Check: The first three terms of the series are \(3/4\), \(-9/64\), and \(243/5120\); the fourth has sign \((-1)^3=-1\), matching the direction of the bound.

A Common Indexing Mistake

A frequent error is to replace \(3^{2k+1}\) by \(3^k\) when matching a sum to the arctangent series. These powers are not interchangeable. For \(k=1\), the arctangent-series term at \(x=1/3\) is \[ -\frac{(1/3)^3}{3}=-\frac{1}{81}, \] not \(-1/9\). The denominator \(2k+1=3\) and the power \(2k+1=3\) both matter. A reliable check is to write out the first two or three terms of the proposed series before identifying its sum.

Another error is to claim more decimal places than the remainder bound supports. A partial sum is not automatically an accurate approximation merely because it has several terms. Compute the next omitted term, form the interval, and check that both endpoints give the same rounding at the requested precision.

Why This Method Is Useful

The arctangent representation turns an infinite sum into a named function value, while the remainder bound makes the representation quantitatively useful. Together, they provide an exact identification and a way to control numerical error without knowing the infinite sum in advance.

The method depends on a precise match: odd powers \(x^{2k+1}\), odd denominators \(2k+1\), alternating signs, and a substitution in the series’ convergence range. Once those features are checked, the partial sums and the next-term bound provide a transparent route from a symbolic series to a certified approximation.

Check Your Understanding

Use the series formula and remainder estimate to answer these questions.

  1. What inverse tangent is represented by \(\sum_{k=0}^{\infty}(-1)^k(1/4)^{2k+1}/(2k+1)\)?
  2. After \(m\) terms at \(x=2/3\), what is the bound on the absolute error?
  3. For a partial sum with \(m=4\) terms and \(x>0\), is the first omitted term positive or negative?
  4. Why is substituting \(x=3/2\) not justified by the arctangent series used here?
  5. What must be checked before claiming that a numerical approximation is correct to a specified number of decimal places?

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