Advanced Calculus Sequences and series: summary

1 Sequences

A sequence {a_{n}} is an infinite list of numbers a_{1}, a_{2}, a_{3} …, a_{n}, . . .. More formally a sequence is a function f : N → R; however we prefer to use the subscript notation a_{n} for the n-th term of the sequence rather than the functional notation f(n).

1.1 Basic properties of sequences

A sequence {a_{n}} is bounded above if there is a number M such that a_{n} ≤ M for every n; such an M is an upper bound for the sequence. Similarly {a_{n}} is bounded below if there is a number m such that m ≤ a_{n} for every n; this m is an lower bound for the sequence. A sequence which is both bounded above and bounded below is called simply a bounded sequence.

A sequence {a_{n}} is increasing if a_{n} < a_{n+1} for each n, and is nondecreasing if a_{n} ≤ a_{n+1} for each n. Similarly {a_{n}} is decreasing if a_{n} > a_{n+1} for each n, and is nonincreasing if a_{n} ≥ a_{n+1} for each n. A sequence that is nondecreasing or nonincreasing is called a monotone sequence.

1.2 Convergence

The notion of convergence is the most important notion in the theory of sequences. Let {a_{n}} be a sequence and A a real number. We say that {a_{n}} converges to A if for all positive real numbers δ there is a number N such that |a_{n} − A| < δ for all n with n ≥ A.

What this means is that if we take any open interval I = (A − δ, A + δ) centred at A then all but finitely many of the terms of the sequence a_{n} lie in I.

If {a_{n}} converges to A we write a_{n} → A (“a_{n} tends to A”) or lim_{n→∞} a_{n} = A (“the limit of a_{n} as n tends to infinity is A”).

A sequence which converges to some real number is called convergent. A sequence which is not convergent is called divergent.

As an example of a convergent sequence define a_{n} = 1/n. This sequence converges to 0. To see that the definition of convergence is satisfied, let δ > 0 and I = (−δ, δ). We need to show that a_{n} ∈ I with only finitely many exceptions. To this end consider the real number 1/δ. Choose an integer N with N > 1/δ. If n ≥ N then n > 1/δ and so 0 < 1/n < δ. Hence a_{n} ∈ I for all n ≥ N, that is a_{n} ∈ I with only finitely many exceptions.

In practice we rarely check convergence by verifying the definition. Instead we use rules whereby the convergence of known sequences imply the convergence of other series.

Constant sequences: If a_{n} = a for all n, then a_{n} → a.

Sum/difference/product/quotient rules: Suppose that {a_{n}} and {b_{n}} are sequences converging to A and B respectively. Then

(a_{n} + b_{n}) → A + B,  (a_{n} − b_{n}) → A − B,  (a_{n}b_{n}) → AB,

and if b_{n} ≠ 0 for all n and B ≠ 0 then

a_{n}/b_{n} → A/B.

Sandwich rule: Suppose that {a_{n}}, {b_{n}} and {c_{n}} are all sequences, and that a_{n} ≤ b_{n} ≤ c_{n} for all n. If a_{n} → A and c_{n} → A (that is they converge to the same limit) then b_{n} → A too.

Bounded monotone sequences: Every bounded monotone sequence is convergent.

Some divergent sequence fail to converge in a fairly disciplined way. We say that a sequence {a_{n}} “diverges to ∞” and we write a_{n} → ∞ if for each real number K there is a number N such that a_{n} > K whenever n ≥ N. That is to say that all but finitely many of the terms of the sequence are bigger than K, no matter what K is. Similarly {a_{n}} “diverges to −∞” (a_{n} → −∞) if for each real number K there is a number N such that a_{n} < K whenever n ≥ N.

It is also useful to remember a few standard limits.

1. a^{n} → 0 if |a| < 1, and a^{n} → ∞ if a > 1,

2. n^{k} → 0 if k < 0, and n^{k} → ∞ if k > 1,

3. \frac{ln n}{n^{k}} → 0 if k > 0,

4. (1 + \frac{a}{n})^{n} → e^{a} for any constant a,

5. n^{1/n} → 1.

2 Series

A series is an “infinite” sum

∑_{n=1}^{∞} = a_{1} + a_{2} + ⋯ + a_{n} + ⋯

of a sequence {a_{n}} of terms. To make sense of this we introduce the partial sums

s_{n} = ∑_{k=1}^{n} a_{k} = a_{1} + a_{2} + ⋯ + a_{n}.

We say that the sequence ∑_{n=1}^{∞} a_{n} converges if the sequence {s_{n}} of its partial sums converges and we write

∑_{n=1}^{∞} a_{n} = lim_{n→∞} s_{n}

if this is the case. Otherwise the series ∑_{n=1}^{∞} a_{n} diverges.

An important example of series is the geometric series given by a_{n} = ar^{n−1}. That is a_{1} = 1 and a_{n+1} = ra_{n} for each n. To consider its convergence we shall compute the partial sums s_{n}. We have

s_{n} = a + ar + ar^{2} + ⋯ + ar^{n−1}

and so

rs_{n} = ar + ar^{2} + ar^{3} + ⋯ + ar^{n}.

Subtracting gives

(1 − r)s_{n} = a − ar^{n}  or  s_{n} = \frac{a(1 − r^{n})}{1 − r}

(at least if r ≠ 1). If |r| < 1 then r^{n} → 0 and by using the difference, product and quotient rules we get a_{n} → a/(1 − r). Hence if |r| < 1 the geometric series is convergent and

∑_{n=1}^{∞} ar^{n−1} = \frac{a}{1 − r} .

2.1 Convergence tests for positive series

If a series ∑_{n=1}^{∞} a_{n} is convergent, then the sequence of its partial sums {s_{n}} is convergent to a number A say. Also s_{n−1} → A and so s_{n} − s_{n−1} → A − A = 0. Hence a_{n} → 0. We get the first of the following convergence/divergence tests.

divergence test: If ∑_{n=1}^{∞} a_{n} converges then a_{n} → 0; equivalently if a_{n} ↛ 0 then ∑_{n=1}^{∞} diverges.

comparison test: Suppose that 0 ≤ a_{n} ≤ b_{n}. If ∑_{n=1}^{∞} b_{n} converges, then ∑_{n=1}^{∞} a_{n} converges. Also if ∑_{n=1}^{∞} a_{n} diverges, then ∑_{n=1}^{∞} b_{n} diverges.

limit comparison test: Suppose that a_{n} > 0 and b_{n} > 0. If (a_{n}/b_{n}) → C where 0 < C < ∞, then the series ∑_{n=1}^{∞} a_{n} and ∑_{n=1}^{∞} b_{n} both converge or diverge.

If (a_{n}/b_{n}) → 0 and ∑_{n=1}^{∞} b_{n} converges then ∑_{n=1}^{∞} a_{n} converges.
If (a_{n}/b_{n}) → ∞ and ∑_{n=1}^{∞} b_{n} diverges then ∑_{n=1}^{∞} a_{n} diverges.

integral test: Let f be a decreasing function defined on the interval [1, ∞); the decreasing condition means that f(x) > f(y) whenever 1 ≤ x < y. If a_{n} = f(n) then the series ∑_{k=1}^{∞} a_{n} is convergent if and only if the improper integral ∫_{1}^{∞} f(x) dx is convergent.

ratio test: Let a_{n} > 0 for each n. Suppose that (a_{n+1}/a_{n}) → k. If k < 1 the series ∑_{n=1}^{∞} a_{n} converges; if k > 1 or if a_{n+1}/a_{n} → ∞ the series ∑_{n=1}^{∞} a_{n} diverges; if k = 1 the test gives no information.

An an example consider the integral test with f(x) = 1/x^{k} for k > 0. Then f is a decreasing function. We need to evaluate the integral I_{k} = ∫_{1}^{∞} dx/x^{k}. For k = 1, I_{1} = [ln x]_{1}^{∞} = ∞ is divergent. Hence the harmonic series ∑_{n=1}^{∞} 1/n diverges. If k ≠ 1 then I_{k} = [x^{1−k}/(1 − k)]_{1}^{∞}. The behaviour of the integral depends on the sign of 1 − k. If 1 − k < 0, that is if k > 1 then x^{1−k} goes to 0 as x goes to infinity. Hence I_{k} = 1/(k − 1) is convergent. But if 1 − k > 0 that is if 0 < k < 1 then x^{1−k} goes to infinity as x goes to infinity and I_{k} = ∞ is divergent. Hence we have that

∑_{n=1}^{∞} \frac{1}{n^{k}} \{ \begin{matrix}is convergent if k > 1, \\ is divergent if 0 < k ≤ 1. \end{matrix}

This result is worth remembering, as the convergence of many series can be tested by comparison to these series.

2.2 Series with mixed signs

If ∑_{n=1}^{∞} |a_{n}| is convergent then so is ∑_{n=1}^{∞} a_{n}. In this case we call the series ∑_{n=1}^{∞} a_{n} absolutely convergent. Given a series ∑_{n=1}^{∞} a_{n} with positive and negative terms, always first test ∑_{n=1}^{∞} |a_{n}| for convergence. If so the original series is convergent.

However the series

∑_{n=1}^{∞} \frac{(−1)^{n−1}}{n} = \frac{1}{1} − \frac{1}{2} + \frac{1}{3} − \frac{1}{4} + ⋯

is convergent despite the series ∑_{n=1}^{∞} 1/n being divergent. Such a series is conditionally convergent. Many series like this with signs alternating between plus and minus are convergent due to the alternating series test or Leibniz’s test: Suppose that {a_{n}} is a sequence such that

1. {a_{n}} is nondecreasing: a_{n} ≥ a_{n+1}, and

2. a_{n} → 0,

then

∑_{n=1}^{∞}(−1)^{n−1}a_{n} = a_{1} − a_{2} + a_{3} − a_{4} + ⋯

is convergent.

2.3 Power series

These are series of the form

∑_{n=0}^{∞} a_{n}x^{n}.

We ask for which values of x these series converge for. Let b_{n} = a_{n}x^{n}. Then

|\frac{b_{n+1}}{b_{n}} | = |\frac{a_{n+1}}{a_{n}} | |x|.

If |a_{n+1}/a_{n}| → L where L is a positive limit, then ∑ a_{n}x^{n} converges when L|x| < 1 and diverges for L|x| > 1, that is the series converges when −R < x < R and diverges for |x| < R where R = 1/L. We call R the radius of convergence of ∑ a_{n}x^{n}.

If |a_{n+1}/a_{n}| → 0 then |\frac{b_{n+1}}{b_{n}} | → 0 too and so the power series ∑ a_{n}x^{n} converges for all x. If |a_{n+1}/a_{n}| → ∞ then the power series diverges for all nonzero x (but converges for x = 0). For all power series ∑_{n=0}^{∞} a_{n}x^{n}, there is a radius of convergence R, which is either a real

number R ≥ 0 or R = ∞ such that either

• R = ∞ and ∑_{n=0}^{∞} a_{n}x^{n} converges for all x,

• R = 0 and ∑_{n=0}^{∞} a_{n}x^{n} converges only for x = 0,

• 0 < R < ∞ and ∑_{n=0}^{∞} a_{n}x^{n} converges when −R < x < R and diverges when |x| > R. (It may converge or diverge when x = ±R.)

Power series can be differentiated and integrated within their radius of convergence. Suppose that

f(x) = ∑_{n=0}^{∞} a_{n}x^{n}

has radius of convergence R. The power series

∑_{n=1}^{∞} na_{n}x^{n−1} = ∑_{m=0}^{∞}(m + 1)a_{m+1}x^{m}

also has radius of convergence R, and equals f′(x) whenever |x| < R. Also the power series

∑_{n=0}^{∞} \frac{a_{n}}{n + 1} x^{n+1} = ∑_{m=1}^{∞} \frac{a_{m−1}}{m} x^{m}

has radius of convergence R and equals

∫_{0}^{x} f(t) dt,

the indefinite integral of f.

RJC 28/2/2001