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NumeriveMATHUniversity mathematics

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Sequences, Convergence & Taylor Series

Useful preparation: Integration Techniques & Improper Integrals

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Sequences track terms; series track running totals. Tiny terms may still accumulate without bound. Taylor series use derivatives at one center, but coefficients, convergence, and equality with the intended function require separate justification.

Sequence and partial sums
A sequence lists terms indexed by positive or nonnegative integers. The partial-sum sequence adds the first finitely many terms of a series.
Absolute and conditional convergence
Absolute convergence means the series of term magnitudes converges. Conditional convergence means the original series converges while that magnitude series diverges.
Taylor polynomial and remainder
The polynomial matches a function's derivatives through a chosen order at a center. The remainder is the actual function value minus this polynomial.

Conditions before calculation

  • Series convergence forces its terms to tend to zero, but that condition alone cannot establish convergence.
  • The alternating-series test requires term magnitudes to tend to zero and be nonincreasing eventually; its next-term error estimate applies after monotonicity begins.
  • A Taylor series represents its function only where the remainders tend to zero. For a power series, test boundary points separately from the open convergence interval.

Why the method works

1. Analyze running totals

Series convergence means its partial sums approach a finite number. This separates term behavior from accumulated behavior.

SN=∑n=1Nan,∑n=1∞an=S  ⟺  lim⁡N→∞SN=SS_N=\sum_{n=1}^{N}a_n,\qquad \sum_{n=1}^{\infty}a_n=S\iff\lim_{N\to\infty}S_N=S

2. Choose a test that reveals structure

Use geometric structure, comparison, or ratios as appropriate. When ratios are eventually defined, a ratio limit of one remains inconclusive.

L=lim⁡n→∞∣an+1an∣:L<1⇒absolute convergence,L>1⇒divergenceL=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|:\quad L<1\Rightarrow\text{absolute convergence},\quad L>1\Rightarrow\text{divergence}

3. Control approximation error

If the next derivative exists and is bounded by M throughout the interval between a and x, Taylor's theorem bounds the error as shown.

PN(x)=∑n=0Nf(n)(a)n!(x−a)n,∣f(x)−PN(x)∣≤M∣x−a∣N+1(N+1)!P_N(x)=\sum_{n=0}^{N}\frac{f^{(n)}(a)}{n!}(x-a)^n,\qquad |f(x)-P_N(x)|\le\frac{M|x-a|^{N+1}}{(N+1)!}

Worked examples

EXAMPLE 1

Sum the geometric series and find its partial-sum error.

  1. The constant ratio has magnitude less than one, so its powers decay.

    SN=∑n=0N(34)n,r=34S_N=\sum_{n=0}^{N}\left(\frac34\right)^n,\qquad r=\frac34

  2. Multiply by the ratio and subtract. All intermediate terms cancel, leaving a finite formula before any infinite operation.

    SN−34SN=1−(34)N+1,SN=4(1−(34)N+1)S_N-\frac34S_N=1-\left(\frac34\right)^{N+1},\qquad S_N=4\left(1-\left(\frac34\right)^{N+1}\right)

  3. Take the limit and subtract the finite sum from it. The error is a remaining tail, not just the next term.

    S=4,S−SN=4(34)N+1S=4,\qquad S-S_N=4\left(\frac34\right)^{N+1}

Conclusion: The sum is 4; every finite cutoff has an explicit positive error.

EXAMPLE 2

Approximate the natural logarithm of 1.2 with three Maclaurin terms and justify accuracy.

  1. Derivatives at zero give the logarithmic coefficients. The expansion is valid inside the stated interval; endpoints require independent tests.

    ln⁡(1+x)=∑n=1∞(−1)n+1xnn,∣x∣<1\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n},\qquad |x|<1

  2. At one fifth, the terms alternate and their magnitudes decrease to zero. Evaluate the first three terms exactly.

    P3(15)=15−150+1375=137750≈0.1826667P_3\left(\frac15\right)=\frac15-\frac1{50}+\frac1{375}=\frac{137}{750}\approx0.1826667

  3. The first omitted term is negative, so this approximation overestimates the value. The alternating-tail estimate bounds the error.

    0<P3(15)−ln⁡(65)<(1/5)44=12500=0.00040<P_3\left(\frac15\right)-\ln\left(\frac65\right)<\frac{(1/5)^4}{4}=\frac1{2500}=0.0004

Conclusion: The three-term estimate has error below four ten-thousandths, with a known direction of error.

Pitfalls & counterexamples

Tempting claim: Terms tending to zero guarantee a convergent series.

Correction: Small terms can still accumulate indefinitely; examine partial sums or apply a justified convergence test.

Test the claim: Reciprocal positive-integer terms tend to zero, but successive doubling blocks in the harmonic series each contribute at least one half.

Check your understanding

If the ratio test gives a limit of one, has the series been shown to diverge?

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Compare your reasoning

No.

The test is inconclusive there: both the harmonic series and the convergent series of reciprocal squares have ratio limit one.

Connect this idea

Taylor polynomials extend tangent-line approximation. Error estimates make infinite-series ideas usable in numerical calculation rather than merely formal expansions.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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