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MATH
[1251]
Taylor series, sequences and infinite series

Infinite series

Define infinite sums through partial sums and recognise geometric, telescoping and harmonic behaviour.

This page gives meaning to an infinite sum

∑n=1∞an=a1+a2+a3+⋯\sum_{n=1}^{\infty}a_n=a_1+a_2+a_3+\cdots

by taking the limit of its finite totals and evaluates the series whose totals have an exact formula. A sequence records individual terms; a series asks whether their accumulated totals settle.

Define an infinite sum through partial sums

The partial sums are finite, so whether they converge is a question about sequences.

[DEFINITION]

Convergence of a sequence

The sequence {an}\{a_n\} converges to LL, written

lim⁡n→∞an=L,\lim_{n\to\infty}a_n=L,

if

∀ε>0 ∃N∈N such that n>N  ⟹  ∣an−L∣<ε.\forall\varepsilon>0\ \exists N\in\mathbb N \text{ such that }n>N\implies |a_n-L|<\varepsilon.
  1. [1]:Seen Questions
  2. [2]:Proving 2+1n+1→22+\frac1{n+1}\to2
  3. [3]:Algebra and continuity of sequence limits
  4. [4]:Squeeze theorem for sequences
[DEFINITION]

Partial sums

For a sequence {an}\{a_n\}, the kk-th partial sum is

sk=∑n=1kan.s_k=\sum_{n=1}^{k}a_n.
  1. [1]:Infinite series
[DEFINITION]

Infinite series

The series ∑n=1∞an\sum_{n=1}^{\infty}a_n converges to LL precisely when its partial sums satisfy

lim⁡k→∞sk=L,sk=∑n=1kan.\lim_{k\to\infty}s_k=L, \qquad s_k=\sum_{n=1}^{k}a_n.

In that case,

∑n=1∞an=L.\sum_{n=1}^{\infty}a_n=L.

The terms and the series answer different questions:

  • ana_n asks what happens to one term far out in the sequence;
  • ∑an\sum a_n asks what happens to the accumulated totals.

A convergent series must have an→0a_n\to0, but this is not sufficient: the harmonic series below is the standard counterexample.

  1. [1]:Partial sums
  2. [2]:Integral test
  3. [3]:Direct comparison test
  4. [4]:Limit comparison test
  5. [5]:Ratio test
  6. 8 more
[PROPOSITION]

Finite changes do not affect convergence

Adding, removing or changing finitely many terms does not affect whether a series converges. If two series converge, their termwise sum and scalar multiples also converge.

Changing finitely many terms changes the partial sums by a fixed amount after the changed terms have been passed. A fixed amount can change the sum, but not whether the partial-sum sequence has a finite limit.

  1. [1]:Infinite series

Series with exact algebraic structure

Some series can be handled without a comparison theorem because their finite partial sums have a clear pattern. Geometric series use a closed formula; telescoping series use cancellation. In both cases, write the finite sum first and take its limit afterwards.

[METHOD]

Evaluate a geometric series

For r≠1r\ne1, first compute the finite partial sum

1+r+⋯+rN=1−rN+11−r.1+r+\cdots+r^N=\frac{1-r^{N+1}}{1-r}.

Then let N→∞N\to\infty. The geometric series converges exactly when ∣r∣<1|r|<1 and in that case

∑n=0∞rn=11−r.\sum_{n=0}^{\infty}r^n=\frac1{1-r}.

If ∣r∣>1|r|>1, the terms do not approach zero. If r=1r=1, the partial sums grow without bound, and if r=−1r=-1, they oscillate.

  1. [1]:Infinite series
  2. [2]:A geometric sum
[EXAMPLE]

A geometric sum

For r=1/2r=1/2,

∑n=1∞12n=12+14+18+⋯=1/21−1/2=1.\sum_{n=1}^{\infty}\frac1{2^n} =\frac12+\frac14+\frac18+\cdots =\frac{1/2}{1-1/2}=1.

[routine]

  1. [1]:Evaluate a geometric series

The geometric formula works because every term is a fixed multiple of the preceding one. Telescoping uses a different kind of repetition: neighbouring terms cancel, leaving only a small boundary contribution.

[METHOD]

Evaluate a telescoping series

If a term can be written as

an=bn−bn+1,a_n=b_n-b_{n+1},

then the finite partial sum is

∑n=0Nan=∑n=0N(bn−bn+1)=b0−bN+1.\sum_{n=0}^{N}a_n =\sum_{n=0}^{N}(b_n-b_{n+1}) =b_0-b_{N+1}.

Take the limit of the single uncancelled tail to decide convergence and find the sum.

Cancelling inside an infinite expression, without first writing the finite sum, can hide a divergent tail.

  1. [1]:Infinite series
  2. [2]:A factorial telescoping series
[EXAMPLE]

A factorial telescoping series

Since

k(k+1)!=1k!−1(k+1)!,\frac{k}{(k+1)!} =\frac1{k!}-\frac1{(k+1)!},

we have

∑k=1Nk(k+1)!=1−1(N+1)!⟶1.\sum_{k=1}^{N}\frac{k}{(k+1)!} =1-\frac1{(N+1)!} \longrightarrow1.

Therefore

∑k=1∞k(k+1)!=1.\sum_{k=1}^{\infty}\frac{k}{(k+1)!}=1.

[exam]

  1. [1]:Evaluate a telescoping series

Why vanishing terms are not enough

For a general series there is no exact partial-sum formula, so the first question is whether the partial sums converge at all. The terms must vanish, but small terms can still accumulate without bound.

[THEOREM]

The nn-th term test

If

lim⁡n→∞an≠0\lim_{n\to\infty}a_n\ne0

or the limit does not exist, then

∑n=1∞an\sum_{n=1}^{\infty}a_n

diverges. Equivalently,

∑an convergent⟹an→0.\sum a_n\text{ convergent}\quad\Longrightarrow\quad a_n\to0.

This is a divergence test, not a convergence test: when an→0a_n\to0, more information is needed. Apply it first, because a nonzero term limit settles the question at once.

  1. [1]:Infinite series
  2. [2]:When the terms do not vanish
  3. [3]:The harmonic series
[EXAMPLE]

When the terms do not vanish

Because

2n3n+1⟶23≠0,\frac{2n}{3n+1}\longrightarrow\frac23\ne0,

the series

∑n=0∞2n3n+1\sum_{n=0}^{\infty}\frac{2n}{3n+1}

diverges.

[routine]

  1. [1]:The nn-th term test
[EXAMPLE]

The harmonic series

Although

1n⟶0,\frac1n\longrightarrow0,

the harmonic series diverges. Group its terms:

1+12+(13+14)+(15+⋯+18)+⋯ .1+\frac12 +\left(\frac13+\frac14\right) +\left(\frac15+\cdots+\frac18\right)+\cdots.

Every block after the first has total at least 1/21/2: the block from 2j−1+12^{j-1}+1 to 2j2^j contains 2j−12^{j-1} terms, each at least 1/2j1/2^j. Hence the partial sums increase by at least 1/21/2 from one block to the next and cannot converge.

[exam]

  1. [1]:The nn-th term test

The contrasting series

∑n=1∞1n2\sum_{n=1}^{\infty}\frac1{n^2}

does converge; for n≥2n\ge2,

1n2<1n(n−1)=1n−1−1n,\frac1{n^2}<\frac1{n(n-1)} =\frac1{n-1}-\frac1n,

so comparison with a telescoping series proves convergence. The exact value π2/6\pi^2/6 is not needed for this convergence argument.

Only special patterns give exact sums. For most series the task is to decide convergence without finding the sum, which is the job of the convergence tests. The same tests later find the radius and endpoints of each power series.

Sequences and limits

Make limits indexed by the integers precise and use them to control approximation and convergence.

Tests for series convergence

Choose and apply the main convergence tests for positive, alternating and power-shaped series.

On this page

Define an infinite sum through partial sumsPartial sumsInfinite seriesFinite changes do not affect convergenceSeries with exact algebraic structureEvaluate a geometric seriesA geometric sumEvaluate a telescoping seriesA factorial telescoping seriesWhy vanishing terms are not enoughThe nn-th term testWhen the terms do not vanishThe harmonic series