Sequences and infinite series

This week you will learn to tell two different objects apart: a sequence is a list of numbers, a series is the sum of the terms of that list. The most common mistake in the exam is to confuse the two. "$a_n\to 0$" is a statement about a sequence; ”$\sum a_n$ converges” is a statement about a series, and the first does not guarantee the second.

Golden rules:

1. A sequence limit is found with the same tools as a function limit: algebra, standard limits, L’Hôpital (switch to the variable $x$ first), squeeze.
2. The sum of a series is the limit of the partial sums: $\sum a_n = \lim s_n$.
3. In telescoping series, write $s_n$ as an explicit formula, then take the limit. “Cancelling by eye” loses points.
4. $a_n = s_n - s_{n-1}$. You can go back from the partial sum to the term.

Sequences

A sequence is a function from the positive integers to the real numbers: $a_1, a_2, a_3,\dots$ If for every $\varepsilon>0$ there is an $N$ such that $\lvert a_n-L\rvert<\varepsilon$ whenever $n>N$, the sequence converges to $L$: $\lim_{n\to\infty}a_n = L$. If there is no such $L$ (or the terms go to $\pm\infty$), the sequence diverges.

Intuition. If the “tail” of the sequence enters an arbitrarily narrow band around $L$ and never leaves it again, the limit is $L$. A finite number of terms at the start does not affect the limit.

Limit rules and tools

The rules for sum, difference, product, quotient (if the limit of the denominator is not zero) and multiplication by a constant are the same as for functions. In addition:

• From a function to a sequence. If $f(x)\to L$ ($x\to\infty$) and $a_n = f(n)$, then $a_n\to L$. This lets you use L’Hôpital. You do not differentiate directly with respect to $n$; write $f(x)$ first.
• Continuous function. If $a_n\to L$ and $g$ is continuous, then $g(a_n)\to g(L)$.
• Squeeze. If $b_n\le a_n\le c_n$ and $b_n, c_n\to L$, then $a_n\to L$. A special case: if $\lvert a_n\rvert\to 0$, then $a_n\to 0$.

Limits to memorize

Growth ordering: $\ln n \ll n^p \ll c^n \ll n! \ll n^n$ ($p>0$, $c>1$). If the denominator of a fraction grows faster, the limit is $0$.

Example 1. $\dfrac{2n^2+1}{5n^2-n}$. Divide by $n^2$: $\dfrac{2+1/n^2}{5-1/n}\to\dfrac25$.

Example 2. $\dfrac{\cos n}{n}$. $\left\lvert\dfrac{\cos n}{n}\right\rvert\le\dfrac1n\to0$, so the limit is $0$.

Example 3. $\dfrac{n}{2^n}$. L’Hôpital for $f(x) = x/2^x$: $\dfrac{1}{2^x\ln 2}\to 0$. The sequence goes to $0$ as well.

Example 4. $\left(1-\dfrac1n\right)^n$. In the standard limit with $x=-1$: $e^{-1}$.

Example 4b. Why is $n^{1/n}\to 1$ true? Write $y = x^{1/x}$, $\ln y = \dfrac{\ln x}{x}$. L’Hôpital: $\dfrac{1/x}{1}\to 0$. So $\ln y\to0$, $y\to e^0=1$. With the same method you can show that sequences like $n^{3/n}$ and $(n^2+1)^{1/n}$ also go to $1$. The pattern is always the same: if the exponent is $1/n$, take $\ln$, you get $\infty/\infty$, apply L’Hôpital.

Sign-alternating sequences

The factor $(-1)^n$ swings the terms between positive and negative. There are two cases:

• If $b_n\to0$, then $(-1)^nb_n\to0$. Reason: $\lvert(-1)^nb_n\rvert = \lvert b_n\rvert\to0$.
• If $b_n\to L\neq0$, then $(-1)^nb_n$ approaches $L$ at even indices and $-L$ at odd indices. There are two different accumulation points, so there is no limit.

The general principle: if two subsequences of a sequence go to different limits, the sequence diverges. $\cos(n\pi) = (-1)^n$ also diverges for the same reason. But $\cos(n\pi)/n\to 0$.

Example 4c. $a_n = \dfrac{(-1)^n n^2}{n^3+1}$. The absolute value $\dfrac{n^2}{n^3+1}\to0$, so $a_n\to0$. Since the degree of the numerator is smaller than that of the denominator, the sign does not matter.

Monotone and bounded sequences

If a sequence grows at every step (or does not shrink), it is monotone increasing (nondecreasing); if it never exceeds some number $M$, it is bounded above. Monotone Sequence Theorem: every monotone and bounded sequence converges. The theorem does not give the value of the limit, it only says that it exists. For recursive sequences you find the value like this:

Example 5. $a_1 = 1$, $a_{n+1} = \dfrac{a_n+6}{2}$. The terms: $1,\ 3.5,\ 4.75,\ 5.375,\dots$ are increasing and do not exceed $6$ (because if $a_n<6$ then $a_{n+1}<6$). If the limit is $L$, then $L = \dfrac{L+6}{2}$, hence $L = 6$.

PITFALL · convergence first. Setting up the equation "$L = f(L)$" assumes that the limit exists. For the sequence $a_{n+1} = 2a_n$, $a_1=1$, the equation gives $L=0$ but the sequence $1,2,4,8,\dots$ diverges. If the exam does not say “assuming it converges”, show also that it is monotone and bounded.

Infinite series

The meaning of the expression $\sum_{n=1}^\infty a_n$ is this. Build the sequence of partial sums:

$s_1 = a_1,\quad s_2 = a_1+a_2,\quad\dots,\quad s_n = \sum_{k=1}^{n}a_k.$

If $s_n\to S$, the series converges and its sum is $S$. If $s_n$ has no limit, the series diverges. So every series question is really a sequence question: the limit of the sequence $s_n$.

From the term to the partial sum, and back. If $s_n$ is given, $a_n = s_n - s_{n-1}$ ($n\ge2$) and $a_1 = s_1$.

Example 5b. The partial sum of a series is given as $s_n = 3-\dfrac{1}{n}$. Then $a_1 = s_1 = 2$, and for $n\ge2$, $a_n = \left(3-\dfrac1n\right)-\left(3-\dfrac{1}{n-1}\right) = \dfrac{1}{n-1}-\dfrac1n = \dfrac{1}{n(n-1)}$. The sum is $\lim s_n = 3$. Note: $a_n\to0$ but the sum is not $0$, it is $3$. The limit of the term and the sum of the series are completely separate things.

Rules for operating on series

If $\sum a_n = A$ and $\sum b_n = B$, then

$\sum(a_n\pm b_n) = A\pm B,\qquad \sum c\,a_n = cA.$

Consequences:

• Convergent + divergent = divergent. ($\sum(\frac1n+\frac1{2^n})$ diverges.)
• Nothing general can be said about divergent + divergent. $\sum 1 + \sum(-1)$ gives $\sum 0 = 0$ term by term.
• Adding or dropping a finite number of terms does not change convergence, it only changes the sum. $\sum_{n=1}^\infty a_n$ and $\sum_{n=10}^\infty a_n$ share the same fate. That is why saying “for large enough $n$” is enough in tests.
• Index shift: $\sum_{n=1}^\infty a_n = \sum_{n=0}^\infty a_{n+1}$. It does not change the sum, only the way it is written.

Telescoping series

If the term can be put in the form $b_n - b_{n+k}$, the middle terms cancel each other in the partial sum.

Example 6. $\displaystyle\sum_{n=1}^\infty\frac{1}{n(n+1)}$. Partial fractions: $\dfrac{1}{n(n+1)} = \dfrac1n-\dfrac{1}{n+1}$.

s_n = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))
    = 1 - 1/(n+1)          (the middle terms cancelled)
lim s_n = 1

Example 7. $\displaystyle\sum_{n=2}^\infty\frac{2}{n^2-1}$. $\dfrac{2}{(n-1)(n+1)} = \dfrac{1}{n-1}-\dfrac{1}{n+1}$. This time the terms cancel two steps apart:

n=2: 1   - 1/3
n=3: 1/2 - 1/4
n=4: 1/3 - 1/5      ← 1/3 cancelled here
n=5: 1/4 - 1/6
...
s_n = 1 + 1/2 - 1/n - 1/(n+1)  →  3/2

The rule for finding how many terms are left: if the difference is $b_n - b_{n+k}$, then $k$ positive terms from the start and $k$ negative terms from the end are left.

TIP · look at the starting index. Example 7 starts at $n=2$. If it started at $n=1$ the term would be undefined ($n^2-1=0$). Always write the remaining terms according to the first index.

The sequence converges, the series diverges

The terms of the harmonic series $\sum\frac1n$ go to $0$, but the sum is infinite. You can see it by grouping:

$1+\tfrac12+\underbrace{\tfrac13+\tfrac14}_{>1/2}+\underbrace{\tfrac15+\dots+\tfrac18}_{>1/2}+\underbrace{\tfrac19+\dots+\tfrac1{16}}_{>1/2}+\dots$

Each group adds at least $1/2$, and there are infinitely many groups. In the same way $\sum\frac{1}{\sqrt n}$ diverges too. The topic of the next two weeks is exactly this: which test tells you whether a series converges when the terms go to zero.

Common mistakes