Basic Analysis: Sequence Convergence (4)

In this article, we’ll consider the convergence of an infinite sum: a_1 + a_2 + a_3 + \ldots. We call this sum an infinite series. Let s_n = a_1 + a_2 + \ldots + a_n be the partial sums of the series.

Definition. We say that \sum_{n=1}^\infty a_i is L (resp. ∞, -∞) if the partial sums (s_n) converge to L (resp. ∞, -∞). The series is said to be convergent if the sum is a real number L.

Example 1.

The series: \frac 1{1\times 2} + \frac 1{2\times 3} + \frac 1{3\times 4} + \ldots converges to 1 since the partial sums

\sum_{i=1}^n \frac 1{i(i+1)} = \sum_{i=1}^n \left(\frac 1 i-\frac 1{i+1}\right) = \left(1 - \frac 1 2\right) + \left(\frac 1 2 - \frac 1 3\right) + \ldots +\left(\frac 1 n-\frac 1 {n+1}\right)=1 - \frac 1{n+1}

form a telescoping series. As n→∞, the RHS clearly converges to 1.

Example 2.

The series: \frac 1 {1^2} + \frac 1 {2^2} + \frac 1{3^2} + \ldots converges. Indeed, we have 0 < \frac 1 {n^2} < \frac 1 {n(n-1)}, so upon dropping the first term (1), we can apply squeeze theorem on the partial sums to obtain:

1 < \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^2} + \ldots < 1 + \frac 1 {1\times 2} + \frac 1 {2\times 3} + \ldots = 2.

Computing the exact value is rather difficult and we won’t go into that now.

Example 3.

Consider the series \sum_{n=1}^\infty \frac 1{4n^2-1}. We can write the sum in two different ways:

\begin{aligned} \sum_n \frac 1{4n^2-1} &= \sum_n \left(\frac n{2n+1} -\frac {n-1}{2n-1}\right) =\frac 1 3 +\left(\frac 2 5 - \frac 1 3\right)+\left(\frac 3 7-\frac 2 5\right) + \ldots\\ &=\sum_n \left(\frac {1/2} {2n-1}-\frac{1/2}{2n+1}\right)= \left(\frac 1 2-\frac 1 6\right)+\left(\frac 1 6-\frac 1 {10}\right) + \ldots \end{aligned}

All the terms in the first sum appear to cancel each other out, leaving a sum of zero, which is ridiculous since all the terms in the series are positive. However, a moment of thought reveals that the partial sums are really \frac n{2n+1} which approaches 1/2. Hence, the series converges to 1/2, which is also obvious from the second sum.

Proposition. If the series \sum a_n is convergent, then (a_n) \to 0.

Proof.

Let ε>0. Since the partial sums converge to a real number, it is a Cauchy sequence so there exists N such that |s_m - s_n| < \epsilon for all mnN. In particular, |a_n| = |s_n-s_{n-1}| < \epsilon for all nN+1. Hence, (a_n) \to 0. ♦

Example 4.

The converse is not true: there’re many examples of sequences (a_n) which converge to 0, but whose sums don’t converge. The most famous example is that of the harmonic series 1 + \frac 1 2 + \frac 1 3 + \ldots, where the sum of the first n terms is approximately log(n).

Basic Properties

Properties. Suppose \sum_n a_n = L_1 and \sum_n b_n = L_2. Then:

  • \sum_n (a_n + b_n) = L_1 + L_2;
  • \sum_n (ca_n) = cL_1 for any real c.

Since the partial sums of (an + bn) is the sum of the partial sum of an and that of bn, the first property follows easily. Similarly, the second property follows from the fact that the partial sums of (c·an) are just (partial sums of an).

On the other hand, the sum of products ∑(anbn) is quite a pain. One way of handling it is to use Abel transformation: write B_n = b_1 + b_2 + \ldots b_n for the partial sums of bn, then:

\begin{aligned}\sum_{n=1}^m a_n b_n &= \sum_{n=1}^m a_n(B_n - B_{n-1}) = \sum_{n=1}^m a_n B_n - \sum_{n=0}^{m-1} a_{n+1} B_n= \sum_{n=1}^m B_n(a_n - a_{n+1})+ a_m B_m.\end{aligned}

Example 5.

Prove that if the partial sums s_n = a_1 + a_2 + \ldots + a_n are bounded, then \frac{a_1}1 + \frac{a_2} 2 + \frac{a_3} 3 + \ldots converges.

Proof.

Let bn = 1/n. By Abel transformation with a‘s and b‘s swapped, we get:

\begin{aligned}\sum_{n=1}^m \frac{a_n} n = \sum_{n=1}^m s_n(b_n - b_{n+1}) + b_m s_m = -\sum_{n=1}^m \frac{s_n}{n(n+1)} + \frac{s_m}m.\end{aligned}

Since (s_n) is bounded, the second term \frac{s_m}m \to 0. Hence it suffices to show that \sum_n \frac {s_n}{n(n+1)} converges. But since |sn|≤L are bounded, we have:

\begin{aligned}\left|\sum_{n=j}^k \frac{s_n}{n(n+1)}\right| \le \sum_{n=j}^k \frac{|s_n|}{n(n+1)} \le L\sum_{n=j}^k \frac 1{n(n+1)}.\end{aligned}

Since \sum_n \frac 1 {n(n+1)} converges, its partial sums form a Cauchy sequence. This in turn implies that the partial sums of \sum_n \frac{s_n}{n(n+1)}  form a Cauchy sequence, and we’re done. ♦

Theorem (Alternating Series Test). If (a_n) is a decreasing sequence of positive values converging to 0, then the series \sum_n (-1)^n a_n is convergent.

Proof.

Let bn = (-1)n. Apply the Abel transformation:

\begin{aligned}\sum_{n=1}^m a_n b_n =\sum_{n=1}^m B_n(a_n - a_{n+1})+ a_m B_m,\end{aligned}

where B_n = b_1 + \ldots + b_n and thus |Bn| ≤ 2. Since a_m \to 0, we also have a_m B_m \to 0 so it suffices to show that \sum_n B_n(a_n-a_{n+1}) converges. Now,

|\sum_{n=j}^k B_n(a_n - a_{n+1})| \overbrace{\le\sum_{n=j}^k |B_n|(a_n-a_{n+1})}^{\because (a_n) \text{ dec.}} \le 2(a_j-a_{k+1}).

Since (a_n) is convergent, it is also Cauchy, so the partial sums of \sum_n B_n(a_n-a_{n+1}) form a Cauchy sequence, and the series is convergent. ♦

Example 6.

Consider the infinite series 1 -\frac 1 2 + \frac 1 3 - \frac 1 4 + \ldots. By the alternating series test, the series converges, though the test doesn’t give any hint what the sum might be.

Permutation of a Series

If \sum_{n\ge 1} a_n is a series, then we can permute the terms to obtain \sum_{n\ge 1} a_{\pi(n)}, where π:N → N is a bijective function. For example:

  • we can switch the first two terms to obtain a_2 + a_1 + a_3 + a_4 + \ldots, i.e. π(1) = 2, π(2) = 1, and π(m) = m for all m > 2;
  • we can switch each 2k-1 with 2k to obtain a_2 + a_1 + a_4 + a_3 + a_6 + a_5 + \ldots, i.e. π(2k-1) = 2k, π(2k) = 2k-1 for k=1, 2, 3, … .

It may surprise the reader that the sum of an infinite series may depend on the order of summation, despite the fact that addition is commutative.

Example 7.

We already saw that S =1 - \frac 1 2 + \frac 1 3 - \ldots is convergent. On the one hand, we can write:

S = 1 - \left(\frac 1 2 - \frac 1 3\right) - \left(\frac 1 4 - \frac 1 5\right) -\ldots

where each bracketed term is positive, so S < 1. On the other hand, let’s write S = 1 + sum of terms of the form:

a_k = \frac 1 {4k-1} + \frac 1 {4k+1} -\frac 1 {2k} = \frac {8k}{16k^2-1} - \frac 1{2k} >0k = 1, 2, 3, … .

This would imply that S > 1, so the sum of an infinite series can be changed by permuting the terms.

For a series to be better-behaved, we need the following concept:

Definition. The series \sum_n a_n is said to converge absolutely if the series \sum_n |a_n| converges. If \sum a_n converges but not absolutely, we say it is conditionally convergent.

For example, the alternating series 1 - \frac 1 2 + \frac 1 3 - \ldots we saw above is only conditionally convergent. On the other hand, the alternating series 1 - \frac 1{2^2} + \frac 1 {3^2}-\ldots is absolutely convergent.

The key property we want to state is:

Theorem. If \sum a_n converges absolutely, then every permutation of the series converges, and to the same sum.

Proof.

Let ε>0. Since the partial sums Σ|an| form a Cauchy sequence, for some N,

|a_m| + |a_{m+1}| + \ldots + |a_n|< \epsilon/2 for all n > mN.

This immediately implies that the partial sums Σan form a Cauchy sequence and converges to some L. Now for any permutation π of {1, 2, 3, … }, we need to show that \sum (a_{\pi(n)} - a_n) = 0. To do that, let:

M = \max\{\pi^{-1}(1), \pi^{-1}(2), \ldots, \pi^{-1}(N)\}.

Basically, the crux is that when nM, π(n) > N. So when n > mM, we have 

\begin{aligned}|\sum_{i=m}^n a_{\pi(i)} - a_i| \le \sum_{i=m}^n |a_{\pi(i)} - a_i| \le \sum_{i=m}^n (|a_{\pi(i)}| + |a_i|) <\frac\epsilon 2 + \frac\epsilon 2.\end{aligned}

So \sum (a_{\pi(n)} - a_n) \to 0 as required. ♦

Hence, absolute convergence of a series ensures well behaviour. This fact carries on over to the “continuous case” where we can exchange integrating variables dx dy and dy dx if the absolute value of the integrand integrates to a finite real value. But that’s another story for another day.

Exercise.

Prove that if \sum_n a_n is conditionally convergent, then for every real L, there is a permutation π of {1, 2, 3, … } such that \sum_n a_{\pi(n)} = L.

[ Hint (highlight to read): let bn and cn be the positive and negative terms in the series (an) respectively. Show that Σbn = ∞ and Σcn = -∞. Now given L, pick just enough terms from bn so that the sum exceeds L. Then pick just enough terms from cn so that the sum drops below L. And so on. Show that bn → 0 and cn → 0 which implies the process eventually tends to L. ]

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