Differentiate, integrate and approximate inside the radius

Try

The geometric identity 1/(1−x)=Σₙ₌₀∞xⁿ holds for |x|<1. Differentiate the finite powers formally and predict a representation for 1/(1−x)². Before trusting the infinite operation, identify the interval where a theorem justifies exchanging differentiation with summation.

Predict

Predict whether differentiating a convergent endpoint series must preserve endpoint convergence. The series Σxⁿ/n at x=−1 converges by alternating signs. Its derivative terms at −1 no longer tend to zero. This provides a counterexample to unrestricted endpoint differentiation.

Experiment

Use geometric differentiation strictly inside |x|<1.

Operation	Result
Differentiate Σₙ₌₀∞xⁿ	Σₙ₌₁∞n x^(n−1)=1/(1−x)²
Multiply by x	Σₙ₌₁∞n xⁿ=x/(1−x)²
Substitute x=1/2	Σn/2ⁿ=2

Compute several partial sums to compare with 2, but use the identity for the exact conclusion. Then integrate 1/(1+t)=Σ(−t)ⁿ from 0 to x to construct the logarithm series, paying attention to the changed indices.

Understand

A power series can be differentiated or integrated term by term inside its radius of convergence, and the resulting power series has the same radius. Uniform convergence on closed subintervals strictly inside the radius underlies the rigorous justification. This is stronger than arbitrary pointwise convergence, which does not generally permit exchanging limits with derivatives or integrals. Full proofs belong to analysis; the open-radius hypothesis remains part of every application here.

Integrating the geometric series for 1/(1+t) gives ln(1+x)=Σₙ₌₁∞(−1)^(n+1)xⁿ/n for |x|<1. The lower endpoint zero fixes the integration constant. At x=1/2, the first three nonzero terms give 1/2−1/8+1/24=5/12. The next alternating term has magnitude 1/64, giving a valid error bound because the magnitudes decrease to zero.

Endpoints require separate work. Differentiating Σxⁿ/n gives Σx^(n−1). At −1 the original alternating harmonic series converges, but the derivative series has terms of magnitude 1 and diverges. An unchanged radius does not mean an unchanged interval with included endpoints. Establishing a limiting function identity at an endpoint may require an additional continuity or Abel-type theorem, not simply substituting into an identity proved only inside.

Series can also provide integrals of functions with inconvenient elementary primitives. The continuous extension of sin t/t at zero has series 1−t²/6+t⁴/120−⋯. Integrating to 1/2 with three terms gives 1/2−1/144+1/19200. The next integrated term has magnitude 1/4515840. The alternating magnitudes decrease on this interval, supplying a justified remainder bound. The integrand’s removable endpoint must be handled rather than called an improper divergence.

Adding and multiplying absolutely convergent power series inside a common radius is justified; products require collecting every contribution to a given power. The coefficient of x² in a product comes from constant-times-quadratic, linear-times-linear and quadratic-times-constant terms. Truncating inputs without checking the desired degree can omit contributions. Conditional endpoint products require more care than this interior rule.

Apply

State the working interval before each operation. Keep indices and constants explicit. For an approximation, report the polynomial value and an independently justified bound. Explain why the endpoint counterexample does not contradict the interior differentiation theorem.

A convergent power series determines its Taylor coefficients

Suppose f(x)=Σ from n=0 to infinity of aₙ(x−c)ⁿ on |x−c|<R, with R>0. The interior termwise-differentiation theorem stated above can be applied repeatedly; each differentiated series still has radius R. After k differentiations, substitute x=c. Terms with n<k have differentiated to zero; terms with n>k contain a positive power of x−c and vanish; the remaining term is k!aₖ. Consequently f⁽ᵏ⁾(c)=k!aₖ and aₖ=f⁽ᵏ⁾(c)/k! for every k. The given series is therefore precisely the Taylor series of its own sum on the open interval. This is a consequence of a supplied positive-radius representation, not a claim that every smooth function equals its Taylor series. Endpoint identities are still separate.

Transfer. Define f by Σ from n=0 to infinity of (3ⁿ/(n+1))·(x−2)ⁿ. Use the ratio test to find its open convergence interval. Find f⁽⁴⁾(2) without first finding a closed form. A learner says the coefficient of (x−2)⁴ is already that derivative; repair the claim. Finally, does the interior argument alone decide the values at the endpoints?

Review the coefficient recovery

The ratio of successive absolute terms tends to 3|x−2|, so R=1/3 and the open interval is (5/3,7/3). The fourth coefficient is 81/5, while the fourth derivative is 4!·81/5=1944/5. The factorial comes from differentiating the fourth power four times. At 7/3 the series becomes the divergent positive harmonic tail; at 5/3 it is an alternating harmonic tail and converges. These are separate endpoint tests, not a conclusion of the interior differentiation theorem. The construction defines f on the open interval; assigning the convergent endpoint sum is an extension and does not by itself establish endpoint differentiability.

Continue

The next lessons change representation from y as a function of x to parameterized and polar curves. Derivative and integral rules continue to apply, but their geometric variables and tracing intervals must be identified carefully.