Choosing a Representation Before Calculating
A series-based evaluation is not just a matter of recalling a familiar formula. The key decisions are which function to represent, what substitution to make, and whether the resulting series can be used in the intended way. A representation that is valid only for certain inputs cannot be substituted into an expression without checking those inputs.
Several useful starting points are the geometric series and the logarithm series. For \(|u|<1\),
Integrating or making a suitable substitution gives, for \(|u|<1\),
Another useful representation, obtained from the geometric series by integration, is the arctangent series. It holds for \(|u|<1\):
At a boundary point, a formula’s behavior must be checked separately; the open-interval conditions above do not automatically include endpoints. In an evaluation problem, the chosen substitution must place the input inside the interval where the representation is valid, or a separate argument must justify the endpoint case.
A Practical Decision Process
Begin by looking for recognizable structure. A denominator of the form \(1-u\) suggests a geometric series. A logarithm or a sum with a \(1/n\) factor may suggest the logarithm series. A sum involving odd powers divided by odd integers may suggest the arctangent series. For a limit with cancellation, look for a series whose first several terms match the quantities being subtracted.
After choosing a representation, write down its condition before substituting. If the series is then integrated, distinguish between validity at each point and the stronger conditions that conveniently allow term-by-term integration. Uniform convergence on the interval is one sufficient condition. It is not necessary: convergence almost everywhere, together with an appropriate domination condition, can also justify passing the limit through the integral.
Choose a known series whose function or algebraic structure appears in the problem.
Determine where the representation is valid, then check the actual substituted values or interval.
For a limit, control the terms left after cancellation. For an integral, establish conditions for exchanging integration and summation.
Compare with the original function, a direct antiderivative, or the behavior of partial sums.
Worked Example: A Limit After Cancellation
Worked Example: A Limit After Cancellation
Evaluate
The logarithm series applies when \(|x|<1\), which includes all \(x\) sufficiently close to zero. Write enough terms to see what remains after cancellation:
Substitute this into the numerator and cancel the matching terms:
After division by \(x^3\), the leading term is \(1/3\). To justify that the remaining terms vanish, fix \(r\) with \(0<r<1\). For \(|x|\le r\), the tail after the cubic term satisfies
Thus the tail divided by \(x^3\) has absolute value at most \(|x|/(1-|x|)\), which tends to zero as \(x\to0\). Therefore the limit is \(1/3\).
Check: The cubic term in the logarithm series is \(x^3/3\), and the subtracted terms remove exactly the linear and quadratic terms. Dividing the remaining leading term by \(x^3\) gives \(1/3\), consistent with the controlled tail estimate.
Worked Example: Integrating Even Though an Endpoint Fails
Worked Example: Integrating Even Though an Endpoint Fails
Evaluate \(\int_0^1 \frac{1}{1+x}\,dx\) using a series. For \(0\le x<1\), substitute \(u=-x\) into the geometric series:
This representation fails to converge at \(x=1\), so it is not valid at every point of the closed integration interval. To justify integration, let \(S_N(x)=\sum_{n=0}^{N}(-1)^n x^n\) be the \(N\)th partial sum. The finite geometric-sum formula gives
For \(0\le x\le1\), the numerator has absolute value at most \(2\), and the denominator is at least \(1\). Hence \(|S_N(x)|\le2\) throughout the interval. For each \(x<1\), \(S_N(x)\) converges to \(1/(1+x)\). The single exceptional point \(x=1\) has measure zero, and the constant function \(2\) is integrable on \([0,1]\). The dominated convergence theorem therefore permits taking the limit of the integrals:
Direct integration evaluates the integral:
Consequently, \(\sum_{n=0}^{\infty}(-1)^n/(n+1)=\ln 2\).
Check: The \(N\)th integrated partial sum is exactly \(\sum_{n=0}^{N}(-1)^n/(n+1)\). The convergence theorem shows these partial sums tend to the integral, and the antiderivative gives \(\ln 2\). The argument does not assume convergence of the original series at the endpoint.
What Justifies Term-by-Term Integration?
The preceding example illustrates an important distinction. A power series may fail to represent a function at an endpoint and still be usable for integration over an interval containing that endpoint. Uniform convergence on the full interval is a convenient sufficient condition for term-by-term integration, but it is not the only possible justification.
More generally, if partial sums converge to the integrand almost everywhere and are bounded in absolute value by an integrable function independent of the partial-sum index, the dominated convergence theorem can justify taking the limit inside the integral. Other convergence theorems may apply under their own hypotheses. The hypotheses must be checked for the particular problem; pointwise equality in the interior alone is not enough to justify exchanging an infinite sum and an integral.
Worked Example: A Definite Integral with Uniform Convergence
Worked Example: A Definite Integral with Uniform Convergence
Evaluate
For \(0\le x\le1/2\), \(0\le x^2\le1/4<1\), so the geometric series gives
This series converges uniformly on the interval: the absolute value of its \(n\)th term is at most \((1/4)^n\), and the bounding geometric series \(\sum_{n=0}^{\infty}(1/4)^n\) converges. Uniform convergence permits term-by-term integration. Multiplying by \(x\) and integrating each term yields
In the fourth line, set \(m=n+1\), so \((1/2)^{2n+2}=(1/4)^m\) and \(2n+2=2m\). The last series is the logarithm series with input \(1/4\), which lies inside its convergence interval.
Check: An antiderivative is \(\frac12\ln(1+x^2)\), since its derivative is \(\frac12\cdot \frac{2x}{1+x^2}=\frac{x}{1+x^2}\). Evaluating at the endpoints gives \(\frac12\ln(5/4)-\frac12\ln(1)=\frac12\ln(5/4)\), agreeing with the series calculation.
Worked Example: An Infinite Sum from the Logarithm Series
Worked Example: An Infinite Sum from the Logarithm Series
Evaluate
The logarithm series represents \(\ln(1+u)\) for \(|u|<1\). The input \(u=2/5\) satisfies this condition, since \(|2/5|<1\). Substituting gives
The first terms provide a useful check on the signs and substitution:
Check: The alternating signs and powers match the expansion of \(\ln(1+u)\), and the input \(2/5\) is strictly within its domain. The evaluated quantity is therefore \(\ln(7/5)\), not \(\ln(2/5)\) or \(\ln(1-2/5)\).
Common Errors to Avoid
Series methods are reliable when each change is accounted for. Three errors are especially easy to make:
- Skipping the domain check. A substitution can move the input outside the interval of convergence. Verify the substituted value or the full range of inputs before using the identity.
- Stopping before cancellation is resolved. In a limit, the first few terms may cancel. Continue until the first nonzero term appears, and control the remaining terms if needed.
- Exchanging an infinite sum and an integral without a reason. State and check a sufficient condition, such as uniform convergence, or use a suitable convergence theorem with its hypotheses established.
The choice of representation is only the start. A complete solution records the domain, makes the required operation legitimate, and checks that the final value agrees with the original problem.
Check Your Understanding
For each question, identify both the appropriate representation and the condition that makes its use legitimate.
- Which series representation is a natural starting point for \(1/(1-3x)\), and what restriction on \(x\) follows?
- In the limit \(\bigl(\ln(1+x)-x\bigr)/x^2\) as \(x\to0\), which term determines the value after cancellation?
- Why does failure of a series at one endpoint not automatically prevent its use in a definite integral?
- What condition on the interval makes the geometric expansion of \(1/(1+x^2)\) uniformly convergent?
- Evaluate \(\sum_{n=1}^{\infty}(-1)^{n+1}(1/3)^n/n\), stating why the logarithm series applies.
