From a Power Series to an Exact Sum
A convergent infinite sum can sometimes be evaluated exactly by recognizing its terms as the coefficients of a known power series. For logarithmic sums, the key pattern is an alternating sign, a power of a fixed number, and a denominator equal to the index.
The logarithm series is especially useful because its coefficients contain \(1/n\). Substituting a suitable value for the variable turns the whole series into a number, and the function on the other side gives that number in closed form. The crucial step is to check that the chosen value lies in the series’ convergence range.
For \(|u|<1\), the series is absolutely convergent: since \(1/n\leq 1\), \(\sum_{n=1}^{\infty}|u|^n/n\leq\sum_{n=1}^{\infty}|u|^n\), and the geometric series on the right converges. This absolute convergence allows substitutions within the stated range and, when needed, addition or subtraction of such sums.
Matching a Sum to the Logarithm Series
Suppose the target sum contains \((-1)^{n+1}r^n/n\). Compare it with the general term \((-1)^{n+1}u^n/n\). If \(u=r\) and \(|r|<1\), the terms match exactly, so the sum is \(\ln(1+r)\). A reliable approach is to identify the variable value from the terms first, check convergence second, and only then substitute.
The sign matters. The first term of this series is \(r\), not \(-r\), which agrees with the first term of \(\ln(1+r)\) near \(r=0\). If the target instead contains \((-1)^n r^n/n\), it is the negative of this series.
Worked Example: A Basic Alternating Reciprocal-Index Sum
Evaluate \(\displaystyle S=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n2^n}\).
Rewrite the power as \((1/2)^n\). The terms then match the logarithm series with \(u=1/2\):
The substitution is valid because \(|1/2|<1\). Therefore,
Check the match by writing the first terms: \(S=1/2-1/8+1/24-\cdots\). These are exactly the logarithm-series terms at \(u=1/2\); the series identity gives their sum as \(\ln(3/2)\). Equivalently, exponentiating the proposed answer gives \(e^S=3/2\), as required by \(S=\ln(1+u)\).
Power Substitutions and Coefficient Matching
A substitution of the form \(u=x^k\), where \(k\) is a positive integer, produces a series involving only powers whose exponents are multiples of \(k\). This is useful when the target terms have powers \(x^{kn}\) rather than \(x^n\).
To justify the substitution, if \(|x|<1\), then \(|x^k|=|x|^k<1\). The logarithm series therefore applies at \(u=x^k\). Substituting gives
This identity also describes the coefficients: the coefficient of \(x^{kn}\) is \((-1)^{n+1}/n\), while powers not divisible by \(k\) do not occur. In other words, the index in the denominator is \(n\), even though the power of \(x\) is \(kn\). Keeping those two roles distinct prevents a common matching error.
The last observation follows by factoring the constant \(1/k\) out of every term. For example, a denominator \(2n\) gives half the sum, while the logarithm’s argument is unchanged.
Worked Example: Terms with Even Powers
Evaluate \(\displaystyle T=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(1/2)^{2n}}{n}\).
The powers are \(x^{2n}\) with \(x=1/2\), so use \(k=2\) in the power-substitution rule. The convergence condition holds because \(|1/2|<1\), and
For a term-by-term check, the first terms are \(1/4-1/32+1/192-\cdots\), matching the logarithm series at \(u=1/4\). The substitution gives \(x^2=1/4\), so the right-hand side is \(\ln(1+1/4)=\ln(5/4)\). The exponent \(2n\) changes the argument to \(1/4\); it does not change the denominator from \(n\) to \(2n\).
Combining Logarithmic Sums
Some sums do not match a single logarithm series at first glance, but can be separated into two matching sums. If both component series converge, their difference is the difference of their sums. The logarithm laws can then simplify that expression.
For example, for \(|a|<1\) and \(|b|<1\), absolute convergence gives
The final equality uses the logarithm law \(\ln A-\ln B=\ln(A/B)\), with \(A=1+a>0\) and \(B=1+b>0\). Positivity follows from \(a,b>-1\), which is guaranteed by \(|a|,|b|<1\).
Worked Example: A Difference of Two Logarithmic Sums
Evaluate \(\displaystyle U=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\left[\left(\frac23\right)^n-\left(\frac13\right)^n\right]\).
Split the expression into two series. Both substitutions are valid because \(|2/3|<1\) and \(|1/3|<1\):
To check the simplification, \((5/3)/(4/3)=(5/3)(3/4)=5/4\). The first terms of the original series are \(1/3-1/6+7/81-\cdots\); the separated series have sums \(\ln(5/3)\) and \(\ln(4/3)\), respectively, so their difference is \(\ln(5/4)\).
Why the Convergence Check Is Essential
The logarithm formula is not valid for every real value of its series variable. In particular, it cannot be substituted at \(u=-1\). Doing so formally would produce \(\ln(0)\) on the function side and the series \(-1-1/2-1/3-\cdots\) on the series side. Neither is a finite real sum: \(\ln(0)\) is undefined, and the harmonic series diverges.
A different endpoint, \(u=1\), is allowed: the series becomes \(1-1/2+1/3-\cdots\), which converges to \(\ln 2\). This is a boundary case, not an instance of the strict condition \(|u|<1\). When a substitution gives an endpoint, check the endpoint result separately rather than assuming the interior formula automatically applies there.
Compare the target term with \((-1)^{n+1}u^n/n\), taking care to distinguish the exponent from the denominator.
Verify \(|u|<1\), or justify a boundary value using the appropriate endpoint result.
Replace the series by \(\ln(1+u)\), then use logarithm laws only when their arguments are positive.
Confirm that the signs, powers, and denominator in the original series agree with the substituted logarithm series.
Common Mistakes
A frequent sign error is to replace \(\sum(-1)^{n+1}u^n/n\) with \(-\ln(1+u)\). For \(u=1/2\), that error gives \(-\ln(3/2)\), even though the first term of the series is \(+1/2\). The correct identity gives \(+\ln(3/2)\). Checking the first term is a quick way to detect a sign reversal.
Another error is to treat \(x^{kn}\) as if its denominator must be \(kn\). The substitution \(u=x^k\) changes the power in the numerator but leaves the index \(n\) in the denominator. If the target has denominator \(kn\), factor out \(1/k\) explicitly. Finally, never use a logarithm-series identity before checking the value substituted into the series.
Why These Sums Matter
These evaluations connect a numerical series to a function with a known value. The connection is useful both for obtaining exact answers and for recognizing which series representation can answer a given question. It also illustrates the wider power-series method: match the coefficients and powers, establish that the substitution is valid, and then use the function represented by the series.
The method is systematic rather than guesswork. A sum containing \((-1)^{n+1}r^n/n\) points to \(\ln(1+r)\); powers such as \(r^{kn}\) point to a substitution of the form \(u=r^k\); and a difference of such terms points to a difference of logarithms. Convergence is part of the reasoning, not an optional final detail.
Check Your Understanding
For each question, identify the logarithm-series substitution and verify that it satisfies the required convergence condition.
- Evaluate \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(1/3)^n}{n}\).
- Evaluate \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(2/5)^{3n}}{n}\).
- How does the value change if the denominator in the second question is \(3n\) instead of \(n\)?
- Why is the logarithm series valid at \(u=1\) but not at \(u=-1\)?
- State the convergence conditions needed to separate a difference of two logarithmic sums.