In this article, we’ll consider the convergence of an infinite sum: . We call this sum an **infinite series**. Let be the **partial sums** of the series.

Definition. We say that is L (resp. ∞, -∞) if the partial sums converge to L (resp. ∞, -∞). The series is said to beconvergentif the sum is a real number L.

**Example 1**.

The series: converges to 1 since the partial sums

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

**Example 2**.

The series: converges. Indeed, we have , so upon dropping the first term (1), we can apply squeeze theorem on the partial sums to obtain:

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

**Example 3**.

Consider the series . We can write the sum in two different ways:

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 which approaches 1/2. Hence, the series converges to 1/2, which is also obvious from the second sum.

Proposition. If the series is convergent, then .

**Proof**.

Let ε>0. Since the partial sums converge to a real number, it is a Cauchy sequence so there exists *N* such that for all *m*, *n* > *N*. In particular, for all *n* > *N*+1. Hence, . ♦

**Example 4**.

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

## Basic Properties

Properties. Suppose and . Then:

- ;
- for any real c.

Since the partial sums of (*a _{n}* +

*b*) is the sum of the partial sum of

_{n}*a*and that of

_{n}*b*, the first property follows easily. Similarly, the second property follows from the fact that the partial sums of (

_{n}*c·a*) are just

_{n}*c·*(partial sums of

*a*).

_{n}On the other hand, the sum of products ∑(*a _{n}*

*b*) is quite a pain. One way of handling it is to use

_{n}**Abel transformation**: write for the partial sums of

*b*, then:

_{n}

**Example 5**.

Prove that if the partial sums are bounded, then converges.

**Proof**.

Let *b _{n}* = 1/

*n*. By Abel transformation with

*a*‘s and

*b*‘s swapped, we get:

Since is bounded, the second term . Hence it suffices to show that converges. But since |*s _{n}*|≤

*L*are bounded, we have:

Since converges, its partial sums form a Cauchy sequence. This in turn implies that the partial sums of form a Cauchy sequence, and we’re done. ♦

Theorem (Alternating Series Test). If is a decreasing sequence of positive values converging to 0, then the series is convergent.

**Proof**.

Let *b _{n}* = (-1)

^{n}. Apply the Abel transformation:

where and thus |*B _{n}*| ≤ 2. Since , we also have so it suffices to show that converges. Now,

Since is convergent, it is also Cauchy, so the partial sums of form a Cauchy sequence, and the series is convergent. ♦

**Example 6**.

Consider the infinite series . 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 is a series, then we can permute the terms to obtain , where π:**N** → **N** is a bijective function. For example:

- we can switch the first two terms to obtain , i.e. π(1) = 2, π(2) = 1, and π(
*m*) =*m*for all*m*> 2; - we can switch each 2
*k*-1 with 2*k*to obtain , i.e. π(2*k*-1) = 2*k*, π(2*k*) = 2*k*-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 is convergent. On the one hand, we can write:

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

, *k* = 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 is said toconverge absolutelyif the series converges. If converges but not absolutely, we say it isconditionally convergent.

For example, the alternating series we saw above is only conditionally convergent. On the other hand, the alternating series is absolutely convergent.

The key property we want to state is:

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

**Proof**.

Let ε>0. Since the partial sums Σ|*a _{n}*| form a Cauchy sequence, for some

*N*,

for all *n* > *m* > *N*.

This immediately implies that the partial sums Σ*a _{n}* form a Cauchy sequence and converges to some

*L*. Now for any permutation π of {1, 2, 3, … }, we need to show that . To do that, let:

Basically, the crux is that when *n* > *M*, π(*n*) > *N*. So when *n* > *m* > *M*, we have

So 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 is conditionally convergent, then for every real *L*, there is a permutation π of {1, 2, 3, … } such that .

[ Hint (highlight to read): let *b _{n}* and

*c*be the positive and negative terms in the series (

_{n}*a*) respectively. Show that Σ

_{n}*b*= ∞ and Σ

_{n}*c*= -∞. Now given

_{n}*L*, pick just enough terms from

*b*so that the sum exceeds

_{n}*L*. Then pick just enough terms from

*c*so that the sum drops below

_{n}*L*. And so on. Show that

*b*→ 0 and

_{n}*c*→ 0 which implies the process eventually tends to

_{n}*L*. ]

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