Infinite Series

An infinite series is a sum

$$a_0 + a_1 + a_2 + \cdots + a_n + \cdots .$$

Notation:

$$\sum_{k=0}^\infty a_k = a_0 + a_1 + a_2 + \cdots + a_n + \cdots .$$

For example,

$$\sum_{k=0}^\infty (-1)^k = 1 - 1 + 1 - 1 + 1 - \cdots + (-1)^n + \cdots,$$

$$\sum_{k=0}^\infty \dfrac{1}{2^k} = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \cdots + \dfrac{1}{2^n} + \cdots,$$

$$\sum_{k=1}^\infty \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} + \cdots.$$

Addition is not defined for an infinite collection of numbers. I have to define what I mean by the sum of an infinite series like those above. To do this, I'll look at the sequence of partial sums. For $a_0 + a_1 + a_2 + \cdots + a_n + \cdots$ , the partial sums are

$$ \eqalign { s_0 &= a_0,\cr s_1 &= a_0 + a_1,\cr s_2 &= a_0 + a_1 + a_2,\cr & \vdots \cr s_n &= a_0 + a_1 + a_2 + \cdots + a_n.\cr } $$

To say that the sum of the series is S means that the sequence of partial sums converges to S:

$$\lim_{n \to \infty} s_n = S.$$

The notation for this is

$$\sum_{k=0}^\infty a_k = S.$$

It is often difficult to compute the sum of an infinite series exactly. However, you can often tell that a series converges without knowing what it converges to. If necessary, a computer can be used to approximate the sum of a convergent series.


Example. The decimal representation of a real number is a convergent infinite series. For example,

$$3.14159265 \ldots 3 + \dfrac{1}{10} + \dfrac{4}{100} + \dfrac{1}{1000} + \dfrac{5}{10000} + \cdots .$$

Repeating decimals represent rational numbers. I'll show by example how to convert a repeating decimal to a rational fraction. Consider $0.272727 \ldots$ . Set $x = 0.272727
   \ldots$ . Then

$$100 x = 27.272727 \ldots, \quad\quad\hbox{so}\quad\quad 99x = 27, \quad\quad\hbox{and}\quad\quad x = \dfrac{27}{99}.\quad\halmos$$


Example. (Geometric series) A geometric series has the form

$$a + ar + ar^2 + ar^3 + \cdots + ar^n + \cdots = \sum_{k=0}^\infty ar^k.$$

The picture below shows the partial sums of the geometric series

$$1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots.$$

Notice that the partial sums seem to approach 2.

$$\hbox{\epsfysize=1.75in \epsffile{series1.eps}}$$

To find a formula for the sum of a geometric series, I'll start by computing the $n^{\rm th}$ partial sum. By long division,

$$\dfrac{1}{1 - r} = 1 + r + r^2 + \cdots + r^n + \dfrac{r^{n+1}}{1 - r}.$$

(This will make sense provided that $r
   \ne 1$ .)

Multiply by a, then move the last term on the right to the left:

$$\sum_{k=0}^n ar^k = a + ar + ar^2 + ar^3 + \cdots + ar^n = \dfrac{a}{1 - r} - \dfrac{ar^{n+1}}{1 - r}.$$

For instance,

$$3 + 3\cdot 5 + 3\cdot 5^2 + \cdots + 3\cdot 5^{100} = \dfrac{3}{1 - 5} - \dfrac{3\cdot 5^{101}}{1 - 5} = \dfrac{3}{4} \left(5^{101} - 1\right).$$

What about the infinite series

$$\sum_{k=0}^\infty ar^k = a + ar + ar^2 + ar^3 + \cdots + ar^n + \cdots?$$

The series converges if the limit of $n^{\rm th}$ partial sum exists. I need to compute

$$\lim_{n \to \infty} \left(\dfrac{a}{1 - r} - \dfrac{ar^{n+1}}{1 - r}\right) = \dfrac{a}{1 - r} - \dfrac{a}{1 - r} \lim_{n \to \infty} r^{n+1}.$$

By a result on geometric sequences,

$$\lim_{n \to \infty} r^{n+1} = \cases{0 & if $|r| < 1$ \cr \hbox{diverges} & if $|r| \ge 1$ \cr}.$$

Hence,

$$\sum_{k=0}^\infty ar^k = \cases{ \dfrac{a}{1 - r} & if $|r| < 1$ \cr \hbox{diverges} & if $|r| \ge 1$ \cr}.$$

For instance,

$$\sum_{k=0}^\infty \dfrac{1}{2^k} = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \cdots + \dfrac{1}{2^n} + \cdots = \dfrac{1}{1 - \dfrac{1}{2}} = 2,$$

$$\sum_{k=0}^\infty 3\cdot 5^k = 3 + 3\cdot 5 + 3\cdot 5^2 + \cdots + 3\cdot 5^n + \cdots\ \hbox{diverges}.$$

What about something like

$$\sum_{k=5}^\infty 4\cdot \left(-\dfrac{1}{3}\right)^k = 4\cdot \left(-\dfrac{1}{3}\right)^5 + 4\cdot \left(-\dfrac{1}{3}\right)^6 + 4\cdot \left(-\dfrac{1}{3}\right)^7 + \cdots ?$$

Well, this is

$$4\cdot \left(-\dfrac{1}{3}\right)^5 \cdot 1 + 4\cdot \left(-\dfrac{1}{3}\right)^5\cdot \left(-\dfrac{1}{3}\right) + 4\cdot \left(-\dfrac{1}{3}\right)^5\cdot \left(-\dfrac{1}{3}\right)^2 + \cdots = \dfrac{4\cdot \left(-\dfrac{1}{3}\right)^5}{1 - \left(-\dfrac{1}{3}\right)} = -\dfrac{1}{81}.$$

How would you use the formula for the infinite series to find, e.g.,

$$\sum_{k=100}^{1000} \dfrac{1}{2^k}?\quad\halmos$$


Example. (Retirement) $200 is deposited each month and collect $4.8\%$ annual interest, compounded monthly. How much is in the account after 30 years?

Note that 30 years is 360 months.

$4.8\%$ annual interest, compounded monthly, means that each month the amount in the account earns $\dfrac{4.8\%}{12} = 0.4\%$ interest. This means that the amount in the account is multiplied by 1.004 each month.

The table below tracks each monthly deposit. The first row represents the first $200 deposited, the second row the second $200 deposited, and so on.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 1 month & & 2 months & & 3 months & & \dots & & 360 months & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 200 & & $1.004\cdot 200$ & & $1.004^2\cdot 200$ & & \dots & & $1.004^{359}\cdot 200$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & 200 & & $1.004\cdot 200$ & & \dots & & $1.004^{358}\cdot 200$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & & & 200 & & \dots & & $1.004^{357}\cdot 200$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & & & & & & & $\vdots$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & & & & & 200 & & $1.004^1\cdot 200$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & & & & & & & 200 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The total amount in the account is the sum of the numbers in the last column, which is

$$200\cdot \sum_{n=0}^{359} 1.004^n = 200\cdot \dfrac{1 - 1.004^{360}}{1 - 1.004} \approx 160429.$$

By comparison, with no interest --- e.g. if you put $200 a month under your mattress --- you'd only have $72000 after 30 years.

At the same time, this is a rather sobering conclusion. Many people would find it a challenge to put away $200 a month toward retirement. This problem shows that doing so and assuming a conservative interest rate produces a significant total --- but hardly enough to retire on!


The following theorem says, roughly, that the sum of two series is the series of their sum, and that constants may be pulled out of series. These are familiar properties of limits, derivatives, and integrals.

Theorem. Suppose that $\displaystyle \sum_{k=1}^\infty a_k$ and $\displaystyle \sum_{k=1}^\infty b_k$ converge, and that c is a constant. Then:

(a) $\displaystyle \sum_{k=1}^\infty
   (a_k + b_k) = \sum_{k=1}^\infty a_k + \sum_{k=1}^\infty b_k$ .

(b) $\displaystyle \sum_{k=1}^\infty
   c\cdot a_k = c\cdot \sum_{k=1}^\infty a_k$ .


Example. Consider the series $\displaystyle \sum_{n=0}^\infty
   \left(\dfrac{1}{4^n} + \left(\dfrac{2}{3}\right)^n\right)$ . This is the sum of two geometric series, both of which converge. Now

$$\sum_{n=0}^\infty \dfrac{1}{4^n} = \dfrac{1}{1 - \dfrac{1}{4}} = \dfrac{4}{3} \quad\hbox{and}\quad \sum_{n=0}^\infty \left(\dfrac{2}{3}\right)^n = \dfrac{1}{1 - \dfrac{2}{3}} = 3.$$

Therefore, $\displaystyle
   \sum_{n=0}^\infty \left(\dfrac{1}{4^n} +
   \left(\dfrac{2}{3}\right)^n\right)$ converges, and

$$\sum_{n=0}^\infty \left(\dfrac{1}{4^n} + \left(\dfrac{2}{3}\right)^n\right) = \dfrac{4}{3} + 3 = \dfrac{13}{3}.\quad\halmos$$


Example. If each of two series diverge, the sum series may still converge. For example, here are two divergent series:

$$\sum_{n=1}^\infty 1 = +\infty \quad\hbox{and}\quad \sum_{n=1}^\infty (-1) = -\infty.$$

But their sum converges:

$$\sum_{n=1}^\infty \left(1 + (-1)\right) = \sum_{n=1}^\infty 0 = 0.$$

Is this happening because "$\infty
   - \infty = 0$ "? No! Here are two divergent series:

$$\sum_{n=1}^\infty n = +\infty \quad\hbox{and}\quad \sum_{n=1}^\infty (-1) = -\infty.$$

Now the sum series diverges:

$$\sum_{n=1}^\infty \left(n + (-1)\right) = \sum_{n=1}^\infty (n - 1) = 0 + 1 + 2 + \cdots = +\infty.$$

No conclusion can be drawn in general about the sum of divergent series.


Example. (Telescoping series) Find

$$\sum_{k=0}^\infty \dfrac{2}{k^2 + 6k + 8}.$$

By partial fractions,

$$\dfrac{2}{k^2 + 6k + 8} = \dfrac{1}{k + 2} - \dfrac{1}{k + 4}.$$

Then

$$\sum_{k=0}^\infty \left(\dfrac{1}{k + 2} - \dfrac{1}{k + 4}\right) = \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) + \left(\dfrac{1}{4} - \dfrac{1}{6}\right) + \left(\dfrac{1}{5} - \dfrac{1}{7}\right) + \left(\dfrac{1}{6} - \dfrac{1}{8}\right) + \cdots .$$

All of the fractions except for $\dfrac{1}{2}$ and $\dfrac{1}{3}$ cancel. Hence,

$$\sum_{k=0}^\infty \dfrac{2}{k^2 + 6k + 8} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}.\quad\halmos$$


In many cases, it's hard to find the exact value of the sum of a convergent series. However, if you know the series converges, you can approximate the sum of the series as closely as you wish by adding up enough terms. Thus, in most of what follows, I'll consider convergence tests, which are methods for determining whether a series converges or diverges (without necessarily finding a sum).

The first test is fundamental: It says that in a convergent series, the terms must go to 0.

Theorem. ( Zero Limit Test) If the series $\displaystyle
   \sum_{k=1}^\infty a_k$ converges, then $\displaystyle \lim_{k\to \infty}
   a_k = 0$ .

Proof. Suppose the series converges to a sum S. Choose some positive number $(\hbox{junk})$ . I'm going to show that if k is large enough, then $a_k$ must lie within $(\hbox{junk})$ of 0.

To do this, find a number n so that $s_k$ and $s_{k-1}$ are within $\dfrac{1}{2}\cdot
   \hbox{junk}$ of S when $k \ge n$ . (I can do this because the series converges to S, so eventually all the terms get arbitrarily close to S.) Then

$$|s_k - S| + |S - s_{k-1}| < \dfrac{1}{2}\cdot \hbox{junk} + \dfrac{1}{2}\cdot \hbox{junk} = \hbox{junk}.$$

But by the Triangle Inequality,

$$|s_k - S| + |S - s_{k-1}| \ge |s_k - S + S - s_{k-1}| = |s_k - s_{k-1}| = |a_k|.$$

Therefore,

$$\hbox{junk} > |a_k| \quad\hbox{for}\quad k \ge n.$$

Now I didn't specify the value of $(\hbox{junk})$ --- it can be {\it any} positive number. So I've just shown that if I take k to be large enough, I can make $a_k$ smaller than any positive number. But this means that the terms must go to 0:

$$\lim_{k \to \infty} a_k = 0.\quad\halmos$$

You usually use the contrapositive of the theorem, which says:

$$\hbox{If}\quad \lim_{k \to \infty} a_k \ne 0, \quad\hbox{then}\quad \sum_{k=1}^\infty a_k \quad\hbox{diverges}.$$


Example. Apply the Zero Limit Test to $\displaystyle \sum_{k=1}^\infty
   \dfrac{5k}{7k + 3}$ .

$$\lim_{k \to \infty} \dfrac{5k}{7k + 3} = \dfrac{5}{7} \ne 0,$$

so the series diverges, by the Zero Limit Test.


Example. Apply the Zero Limit Test to $\displaystyle \sum_{k=1}^\infty \cos k$ .

$$\lim_{k \to \infty} \cos k \quad\hbox{does not exist},$$

so the series diverges, by the Zero Limit Test.


Warning: A standard mistake is to use the Zero Limit Test backward. It is not true that if $\displaystyle \lim_{k \to \infty} a_k = 0$ , then the series $\displaystyle \sum_{k=1}^\infty a_k$ converges. A counterexample, the harmonic series $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k}$ , is discussed in the next example.


Example (The harmonic series) The series

$$\sum_{k=1}^\infty \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} + \cdots$$

is called the harmonic series.

Draw the graph of $y = \dfrac{1}{x}$ , $1
   \le x \le n + 1$ . Divide the interval $1 \le x \le n + 1$ into $n + 1$ equal subintervals of length 1, and build a rectangle on each subinterval using the left-hand endpoints for the heights.

$$\hbox{\epsfysize=2in \epsffile{series2.eps}}$$

The areas of the rectangle are 1, $\dfrac{1}{2}$ , ..., $\dfrac{1}{n}$ . Their sum is

$$s_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n},$$

which is the n-th partial sum of the harmonic series.

Since the rectangles lie above the curve,

$$s_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} > \int_1^{n+1} \dfrac{1}{x}\,dx = \left[\ln x\right]_1^{n+1} = \ln (n + 1).$$

Take the limit as $n \to \infty$ :

$$\lim_{n \to \infty} s_n > \lim_{n \to \infty} \ln (n + 1) = +\infty.$$

Therefore, $\lim_{n \to \infty} s_n =
   +\infty$ : The harmonic series {\it diverges}.

By comparison,

$$\sum_{k=1}^\infty \dfrac{1}{k^2} = \dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \cdots = \dfrac{\pi^2}{6}.$$

I'll have more to say about series of the form $\sum_{k=1}^\infty \dfrac{1}{k^p}$ below.

Finally, go back to the inequality above. I have

$$1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} > \ln (n + 1) > \ln n.$$

Look at

$$a_n = \left(1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n}\right) - \ln n.$$

It's possible to show that $\lim_{n \to
   \infty} a_n$ converges. The limit is denoted $\gamma$ , and is called Euler's constant:

$$\gamma = \lim_{n \to \infty} \left(1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} - \ln n\right).$$

$\gamma \approx 0.5772156649 \ldots$ . So, for example, the sum of the first one million terms of the harmonic series is

$$1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{10^6} \approx \ln 10^6 + \gamma \approx 13.23830.$$

Question: Is $\gamma$ irrational? The answer is not known.


Here are a couple of neat points about the last example.

First, the method I used --- that of comparing a series to an integral --- can be used on series of the form

$$\sum_{k=1}^\infty \dfrac{1}{k^p}, \quad p > 0.$$

These are called p-series; the harmonic series is the case $p = 1$ . A p-series converges if $p > 1$ and diverges if $0 < p \le 1$ . For example,

$$\sum_{k=1}^\infty \dfrac{1}{k^3} \hbox{ converges},$$

$$\sum_{k=1}^\infty \dfrac{1}{\sqrt{k}} \hbox{ diverges},$$

$$\sum_{k=1}^\infty k^{-3/4} \hbox{ diverges}.$$

If $p > 1$ , the sum of the p-series is denoted $\zeta(p)$ . Thus,

$$\zeta(3) = \sum_{k=1}^\infty \dfrac{1}{k^3}.$$

It isn't too difficult to find closed form expressions for $\zeta(2n)$ , where n is an integer. For instance,

$$\zeta(2) = \sum_{k=1}^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{6}.$$

However, the odd sums $\zeta(2n + 1)$ are somewhat mysterious. It was only in 1978 that R. Ap\'ery showed that $\zeta(3)$ is irrational. No one knows what its exact value is, and no one knows if (for instance) $\zeta(5)$ is irrational.

Second, the method of comparing a series to an integral works more generally.

Theorem. Suppose $\displaystyle \sum_{k=1}^\infty a_k$ is a series in which the terms are positive. Let $f(x)$ be the function you get by replacing k by x in $a_k$ . Suppose that:

(a) f is continuous for $x \ge 1$ .

(b) f decreases for $x \ge 1$ .

Then:

$$ \eqalign { \hbox{If}\quad \int_1^\infty f(x)\,dx \quad\hbox{converges, so does}\quad & \sum_{k=1}^\infty a_k.\cr \hbox{If}\quad \int_1^\infty f(x)\,dx \quad\hbox{diverges, so does}\quad & \sum_{k=1}^\infty a_k. \quad\halmos \cr } $$

This is called the Integral Test. Here's why it works. Divide the interval $[1, n +
   1]$ up into rectangles of width 1, using the left-hand endpoints to get the heights. The function f decreases, so the picture looks like this:

$$\hbox{\epsfysize=2in \epsffile{series3.eps}}$$

The sum of the rectangle areas is the $n^{\rm th}$ partial sum, and it is clearly bigger than the area under the curve:

$$s_n = f(1) + f(2) + \cdots + f(n) > \int_1^{n+1} f(x)\,dx.$$

Therefore, if $\displaystyle \lim_{n
   \to \infty} \int_1^{n+1} f(x)\,dx$ diverges, so does $\displaystyle \lim_{n \to \infty} s_n$ , but $\displaystyle
   \lim_{n \to \infty} s_n$ is the sum of the series.

Next, divide the interval $[1, n]$ up into rectangles of width 1, but use the right-hand endpoints to get the heights. The picture looks like this:

$$\hbox{\epsfysize=2in \epsffile{series4.eps}}$$

The sum of the rectangle areas is clearly smaller than the area under the curve:

$$f(2) + f(3) + \cdots + f(n) < \int_1^n f(x)\,dx.$$

If I add $f(1)$ to both sides, the left side becomes the n-th partial sum:

$$s_n = f(1) + f(2) + f(3) + \cdots + f(n) < f(1) + \int_1^n f(x)\,dx.$$

If $\displaystyle \lim_{n \to \infty}
   \int_1^n f(x)\,dx$ converges, so does $\displaystyle \lim_{n \to \infty}
   s_n$ , but $\displaystyle \lim_{n \to \infty} s_n$ is the sum of the series.


Example. Think of using the Integral Test when the general term of the series looks like something that you can integrate.

Consider the series $\displaystyle
   \sum_{k=1}^\infty ke^{-k}$ . $xe^{-x}$ is something you can integrate, so it's natural to try the Integral Test.

For $k \ge 1$ , $ke^{-k} > 0$ . The series has positive terms.

Let $f(x) = xe^{-x}$ . f is continuous for $x \ge 1$ . Since

$$f'(x) = (1 - x)e^{-x} < 0 \quad\hbox{for}\quad x > 1,$$

f decreases for $x \ge 1$ . The hypotheses of the Integral Test are satisfied. (It's important to check the hypothesis before applying the test!)

Compute the improper integral:

$$\int_1^\infty xe^{-x}\,dx = \lim_{c \to \infty} \int_1^c xe^{-x}\,dx = \lim_{c \to \infty} \left[-xe^{-x} - e^{-x}\right]_1^c =$$

$$\lim_{c \to \infty} \left(-ce^{-c} - e^{-c} + e^{-1} + e^{-1}\right) = 2e^{-1}.$$

Here's the parts table:

$$\matrix{& \der {} x & & \displaystyle \int\,dx \cr & & & \cr + & x & & e^{-x} \cr & & \searrow & \cr - & 1 & & -e^{-x} \cr & & \searrow & \cr + & 0 & \rightarrow & e^{-x} \cr}$$

Since the integral converges, the series converges by the Integral Test.


Warning: The value of the integral in the Integral Test is not equal to the sum of the series.

The proof of the Integral Test yields the formula

$$f(1) + \int_1^n f(x)\,dx > s_n > \int_1^{n+1} f(x)\,dx.$$

You can use this to estimate the partial sums of a series to which the Integral Test applies.


Example. Estimate the sum of the first 1000 terms of

$$\sum_{k=1}^\infty \dfrac{1}{{\root 3 \of k}}.$$

This is a divergent p-series with $p =
   \dfrac{1}{3}$ , so the Integral Test applies. I have

$$f(1) + \int_1^{1000} \dfrac{1}{x^{1/3}}\,dx > s_{1000} > \int_1^{1001} \dfrac{1}{x^{1/3}}\,dx.$$

Now

$$f(1) + \int_1^{1000} \dfrac{1}{x^{1/3}}\,dx \approx 149.5 \quad\hbox{while}\quad \int_1^{1001} \dfrac{1}{x^{1/3}}\,dx \approx 148.59998.$$

Thus, $148.59998 < s_{1000} < 149.5$ .


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