Integrating a Maclaurin Series and Retesting Endpoints

You integrate a Maclaurin series term-by-term and find the new series has the exact same radius of convergence as the original, but the endpoints, where your old series blew up, may now converge. That is the central puzzle of integrating a Maclaurin series and retesting endpoints. The reason is mechanical: integration inserts a factor of 1/(n+1) into each term, which changes the tail of the series from a divergent harmonic progression into a convergent p-series at one boundary but not the other.

The radius of convergence R comes from the ratio of coefficients only, the a_n values in the original sum ∑ a_n x^n. When you integrate, the new coefficients become a_n/(n+1). The ratio test limit for the integrated series uses lim | (a_{n+1}/(n+2)) / (a_n/(n+1)) |. The (n+1) and (n+2) cancel in the limit, leaving the same limit as the original series. So R does not budge.

Endpoints are different because the ratio test is inconclusive there. At x = c ± R, you cannot rely on coefficient ratios; you must plug the endpoint into the new integrated series and apply a separate convergence test. Integration changes the exponent of n in the denominator, converting a p=1 harmonic series (which diverges) into a p>1 series (which converges) or vice versa. That is why the radius stays fixed but endpoint convergence can flip.

Start with the geometric series 1/(1-x) = ∑ x^n from n=0 to ∞, which is a Maclaurin series centered at 0. Its radius of convergence is R=1, found by the ratio test: lim |a_n/a_{n+1}| = lim 1/1 = 1. At x=1 the series becomes ∑ 1, which diverges by the nth-term test. At x=-1 it becomes ∑(-1)^n, which also diverges because terms do not approach zero. The interval of convergence for the original series is (-1,1), open at both ends.

Now integrate term-by-term: ∫ 1/(1-x) dx = -ln(1-x) = ∑ x^{n+1}/(n+1). Re-index with k=n+1, giving ∑ x^k/k from k=1 to ∞. This integrated series also has R=1 because the factor 1/k in the coefficient does not affect the radius.

At x=1 the integrated series becomes ∑ 1/k, the harmonic series, which diverges. At x=-1 it becomes ∑ (-1)^k/k, the alternating harmonic series, which converges by the alternating series test. The integrated series therefore converges on [-1,1): one endpoint now works. The radius stayed 1, but the interval gained the left endpoint.

To confirm the radius, run the ratio test on the integrated series ∑ a_k x^k with a_k = 1/k. The ratio limit is lim |a_{k+1}/a_k| = lim (1/(k+1)) / (1/k) = lim k/(k+1) = 1, so R = 1/1 = 1. The radius matches the original. For a quick check of your own work, a maclaurin polynomial calculator will show the first several terms and their coefficients, but it cannot test endpoints, that requires the p-series test and alternating series test by hand.

At x=1 the series is ∑ 1/k. This is a p-series with p=1, which diverges. At x=-1 the series is ∑ (-1)^k/k. Terms decrease to zero, so the alternating series test guarantees convergence. The convergence is conditional: ∑ |(-1)^k/k| = ∑ 1/k diverges. That conditional convergence is only possible at endpoints.

Term-by-term integration cannot change the radius because the ratio test operates on the leading coefficient ratio, which the factor 1/(n+1) does not alter in the limit. But at the endpoints the ratio test is powerless. You must substitute the endpoint into the new series and identify its type. The integrated series often becomes a p-series or an alternating series, and those tests decide convergence.

The 1/(n+1) factor can turn a divergent p=1 series into a convergent p=2 series at one endpoint if the integration added an extra power to the denominator. In the example above, the original series had no denominator at all (∑ x^n), so at x=1 it was ∑ 1, which diverges. The integrated series ∑ 1/k has a denominator, but only p=1 still diverges. If the original series had been ∑ x^n / n, the integrated series would have ∑ x^n / (n(n+1)) at one endpoint, which is a p=2 series and converges.

Integration can also remove convergence. If the original Maclaurin series converges at an endpoint, the integrated series may diverge there. For example, the Maclaurin series for ln(1+x) = ∑ (-1)^{n-1} x^n / n converges at x=1 (alternating harmonic series). Integrate term-by-term: ∫ ln(1+x) dx = (1+x)ln(1+x) - x = ∑ (-1)^{n-1} x^{n+1} / [n(n+1)]. At x=1 this becomes ∑ (-1)^{n-1} / [n(n+1)]. The absolute value ∑ 1/[n(n+1)] converges by comparison to ∑ 1/n^2, so the integrated series converges. But consider a series that converges conditionally at an endpoint, integration may weaken the denominator just enough to produce divergence. This is a known failure case: if the original series converges conditionally at an endpoint, term-by-term integration can cause divergence there.

Students find the radius, declare the interval open, and never plug the endpoints into the integrated series. The ratio test is silent at boundaries. If you stop after computing R, you will miss the endpoint that integration just made convergent. Always retest both endpoints separately after term-by-term integration, using a p-series or alternating series test on the simplified endpoint series. That is the single step that separates a correct interval from a lost point.