Geometric series, the nth-term test and the integral test

This week we give the first three answers to the convergence question. The geometric series is one of the rare series whose sum you can find with a closed formula; in the exam it almost always comes as a “find the sum” question. The nth-term test (divergence test) is the fastest divergence check. The integral test ties the fate of a series to an improper integral and proves the p-series rule. Since next week’s comparison tests all work by comparing to p-series and geometric series, the three results on this page are the cornerstone.

Golden rules:

1. Geometric series: sum $=\dfrac{\text{first term}}{1-r}$, only when $\lvert r\rvert<1$.
2. In every series, first look at whether the term goes to zero. If it does not, you are done: it diverges.
3. If the term goes to zero, the nth-term test says nothing. Move on to another test.
4. The p-series $\sum 1/n^p$: converges for $p>1$, diverges for $p\le1$.

Geometric series

Definition. A series in which each term is a fixed multiple $r$ of the previous one:

$\sum_{n=0}^{\infty}ar^n = a + ar + ar^2 + \dots,\qquad a\neq0.$

$r$ is called the common ratio.

Theorem. If $\lvert r\rvert<1$ the series converges and its sum is $\dfrac{a}{1-r}$. If $\lvert r\rvert\ge1$ it diverges.

Why? The partial sum: $s_n = a+ar+\dots+ar^{n-1}$. Taking the difference with $rs_n$, almost everything cancels: $s_n - rs_n = a - ar^n$, so

$s_n = \frac{a(1-r^n)}{1-r}\qquad(r\neq1).$

For $\lvert r\rvert<1$, $r^n\to0$ and $s_n\to\dfrac{a}{1-r}$. For $\lvert r\rvert>1$, $r^n$ blows up. For $r=1$, $s_n=na$; for $r=-1$, $s_n$ goes back and forth between $a$ and $0$. Both diverge.

Recognizing a geometric series in the exam

The series is usually not given in the form $ar^n$. You separate the exponents and rearrange:

1) Put the term in the form c · (number)^n:  2^(n+1)/3^n = 2 · (2/3)^n
2) The ratio r = (number).  Is |r| < 1?
3) First term = the term at the starting index (plug in n=0, 1, 2 ... whichever it is)
4) Sum = first term / (1 - r)

Example 1. $\displaystyle\sum_{n=1}^{\infty}5\left(\frac13\right)^{n-1}$. The first term at $n=1$ is $5$. The ratio is $\frac13$. The sum is $\dfrac{5}{2/3} = \dfrac{15}{2}$.

Example 2. $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}$. Rearrange: $\dfrac13\left(-\dfrac23\right)^n$. The first term is $\frac13$, the ratio is $-\frac23$. The sum is $\dfrac{1/3}{1+2/3} = \dfrac{1/3}{5/3} = \dfrac15$.

Example 3 (repeating decimal). $1.2\overline{34} = 1.2343434\ldots$ First separate the non-repeating part: $1.2 + 0.0343434\ldots$ The second part is a geometric series with first term $\dfrac{34}{1000}$ and ratio $\dfrac{1}{100}$:

$\frac{34/1000}{1-1/100} = \frac{34}{990}.\qquad 1.2+\frac{34}{990} = \frac{1188+34}{990} = \frac{1222}{990} = \frac{611}{495}.$

Example 3b (application). A ball is dropped from a height of $10$ m. On each impact it bounces to $\frac34$ of the previous height. How far does the ball travel in total? The first drop is $10$ m. After that, each bounce goes both up and down: $2\left(10\cdot\frac34+10\cdot\left(\frac34\right)^2+\dots\right)$. The expression in parentheses is a geometric series with first term $7.5$ and ratio $\frac34$, with sum $\dfrac{7.5}{1/4} = 30$. The total distance is $10+2\cdot30 = 70$ m. Do not forget to count the first drop separately: it goes one way only.

Example 3c (a series with x). For which $x$ does $\sum_{n=0}^\infty (3x)^n$ converge? The ratio is $3x$, the condition is $\lvert3x\rvert<1$, that is, $-\frac13<x<\frac13$. In this interval the sum is $\dfrac{1}{1-3x}$. This idea is the starting point of the power series in week 5: writing a function like an infinite polynomial.

PITFALL · the first term. The $a$ in the formula is not the “coefficient”, it is the first term. In the series $\sum_{n=2}^\infty(1/2)^n$, $a = 1/4$, and the answer is $\dfrac{1/4}{1/2} = \dfrac12$. Taking the coefficient as $1$ and writing $2$ is the most common mistake.

The nth-term test (divergence test)

Theorem. If $\sum a_n$ converges, then $a_n\to0$.

The form you actually use is the contrapositive: if $\lim a_n\neq0$ or the limit does not exist, then $\sum a_n$ diverges.

The reason is simple: $a_n = s_n - s_{n-1}$. If the series converges to $S$, both $s_n$ and $s_{n-1}$ go to $S$, and the difference goes to $0$.

Example 4. $\sum\dfrac{n}{3n-1}$: the term $\to\frac13\neq0$, diverges.

Example 5. $\sum(-1)^{n}$: the term has no limit, diverges.

Example 6. $\sum n\sin(1/n)$: $n\sin(1/n) = \dfrac{\sin(1/n)}{1/n}\to1$, diverges.

PITFALL · the converse is not true. There is NO rule that says if $a_n\to0$ then the series converges. For $\sum1/n$ (the harmonic series) the terms go to zero but it diverges. The nth-term test only proves divergence; an answer that says “converges by the nth-term test” gets zero.

The integral test

Theorem. Let $a_n = f(n)$ and let $f$ be positive, continuous and decreasing for $x\ge N$. Then

$\sum_{n=N}^\infty a_n\quad\text{and}\quad\int_N^\infty f(x)\,dx$

either both converge or both diverge.

Intuition. Draw the values $a_n$ as rectangles of width $1$. Since $f$ is decreasing, the rectangles can be arranged to stay under the curve (right endpoint) or to stick out above it (left endpoint):

$\int_{1}^{n+1}f\,dx\;\le\;s_n\;\le\;a_1+\int_1^n f\,dx.$

If the area is finite, the sum stays bounded (positive terms, increasing partial sums, so it converges). If the area is infinite, the sum is infinite too.

Do not apply the test without writing the three conditions. In the exam, write the sentence “f is positive, continuous and decreasing for $x\ge1$”. The shortest way to show that it is decreasing is $f'(x)<0$.

Example 7. $\displaystyle\sum_{n=1}^\infty\frac{1}{1+n^2}$. $f(x) = \dfrac{1}{1+x^2}$ is positive, continuous, decreasing ($f' = \dfrac{-2x}{(1+x^2)^2}<0$).

$\int_1^\infty\frac{dx}{1+x^2} = \lim_{b\to\infty}\arctan b - \arctan1 = \frac\pi2-\frac\pi4 = \frac\pi4.$

It is finite, so the series converges. The sum is NOT $\pi/4$; the integral only tells the fate. (As a bound: sum $\le a_1+\pi/4 = \frac12+\frac\pi4$.)

Example 7b. $\displaystyle\sum_{n=1}^\infty\frac{n}{n^2+1}$. The term goes to zero, and the nth-term test says nothing. $f(x) = \dfrac{x}{x^2+1}$ is positive and continuous. $f'(x) = \dfrac{1-x^2}{(x^2+1)^2}$ is negative for $x>1$, so it is decreasing after $x\ge1$. With the substitution $u = x^2+1$

$\int_1^b\frac{x\,dx}{x^2+1} = \frac12\ln(b^2+1)-\frac12\ln2\to\infty.$

The integral diverges, so the series diverges. Intuition: for large $n$ the term is $\approx\frac1n$, it behaves like the harmonic series. This “look at the dominant term” intuition is the very idea of next week’s limit comparison test.

TIP · the decrease can start later. In the integral test, $f$ does not have to be decreasing from the very beginning. “Decreasing for $x\ge N$” is enough, because a finite number of terms does not change the fate. In Example 7b, $f$ actually peaks at $x=1$; starting the test at $N=1$ is not a problem.

p-series

For $\sum_{n=1}^\infty\dfrac{1}{n^p}$, $f(x) = x^{-p}$ ($p>0$) is positive, continuous, decreasing. The p-integral result from week 1 applies directly:

Examples: $\sum\frac{1}{n^2}$ converges. $\sum\frac{1}{\sqrt n}$ ($p=\frac12$) diverges. $\sum\frac{1}{n^{1.01}}$ converges (very slowly, but it converges). $\sum\frac1n$ diverges. If $p\le0$, the terms do not even go to zero.

The logarithmic family

$\sum\dfrac{1}{n(\ln n)^p}$ ($n\ge2$): with the substitution $u=\ln x$ the integral becomes $\int\frac{du}{u^p}$. The rule is the same: converges for $p>1$, diverges for $p\le1$. So $\sum\frac{1}{n\ln n}$ diverges, $\sum\frac{1}{n(\ln n)^2}$ converges.

Remainder estimate

If $S$ is the true sum, $s_n$ the partial sum and $R_n = S-s_n$, then under the conditions of the integral test:

$\int_{n+1}^\infty f(x)\,dx\;\le\;R_n\;\le\;\int_n^\infty f(x)\,dx.$

Example 8. If you use $s_{100}$ for $\sum 1/n^2$, the error is at most $\int_{100}^\infty x^{-2}dx = 0.01$. If it says “let the error be less than $0.0001$”, then $\frac1n\le0.0001$, that is, $n\ge10000$ terms are needed.

Common mistakes