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A series converges when its partial sums do, which is a statement about a different list of numbers from the terms; and the geometric series, where $\dfrac{a}{1-r}$ gives the sum outright provided $|r| < 1$ and $a$ is the first term actually there.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can tell a sequence from the series built out of it, and say which of the two any question is about. You can build partial sums and use them as the definition of convergence, recognise a geometric series and check the hypothesis $|r| < 1$ before quoting $\dfrac{a}{1-r}$, identify the first term actually present when a series starts late, and sum a telescoping series by writing its partial sum down.
You have summed a finite geometric series in algebra, and you have met improper integrals, where an infinite region sometimes has a finite area and sometimes does not. A series asks that same question about an infinite sum, and the answer turns on the same thing: how fast the pieces shrink.
Sequence: a list of numbers $a_1, a_2, a_3, \ldots$
Series: what you get by adding a sequence up, written $\sum a_n$.
Partial sum $s_N$: the total of the first $N$ terms — a finite number, and there is one for every $N$.
Converges: the partial sums approach a finite limit. That limit is called the sum of the series.
Geometric: each term is a fixed multiple $r$ of the one before it.
Adding infinitely many numbers is not an operation you can perform, so the definition sidesteps it. Build the partial sums
$$s_1 = a_1,\quad s_2 = a_1 + a_2,\quad s_3 = a_1 + a_2 + a_3,\ \ldots$$
Each is an ordinary finite total. The series converges when this new sequence $s_1, s_2, s_3, \ldots$ approaches a limit, and that limit is what $\sum a_n$ means. Otherwise the series diverges.
So there are two sequences in play at all times: the terms and the partial sums. Every statement in this unit is about one or the other, and mixing them up is the mistake the whole unit is built to prevent.
The geometric series is the case you can finish. If each term is $r$ times the one before, then
$$a + ar + ar^2 + \cdots = \frac{a}{1 - r}\qquad\text{provided } |r| < 1,$$
and it diverges otherwise. Two things in that sentence are easy to lose. The condition $|r| < 1$ is a hypothesis, not a footnote: without it the formula returns a number that is not the sum of anything. And $a$ is the first term actually present — a series written from $n = 3$ has the same $r$ and a different $a$.
Telescoping is the other case where $s_N$ can be written down. If $a_n = b_n - b_{n+1}$ then almost everything cancels and $s_N = b_1 - b_{N+1}$, so the sum is whatever $b_1 - \lim b_{N+1}$ comes to.
Another way: picture
Picture walking toward a wall, each step half of what is left. Your step lengths are the terms and they go to zero; your position is the partial sum and it approaches the wall. Now picture steps of $1, \tfrac12, \tfrac13, \tfrac14, \ldots$ metres: the steps still shrink to nothing, and you walk past every wall there is. Shrinking steps and a finite journey are not the same condition.
Another way: steps
| The question | Which list | What the answer means |
|---|---|---|
| Does $a_n$ converge? | the terms | they settle to some number |
| Does $\sum a_n$ converge? | the partial sums | the running total settles |
| What is $\lim a_n$? | the terms | the size of a late term |
| What is $\sum a_n$? | the partial sums | the total of all of them |
The harmonic series answers the first question with yes, to zero and the second with no, at the same time and with no contradiction. Terms going to zero is necessary and nowhere near sufficient. The harmonic series has terms going to zero and its sum is infinite; that single example is why the $n$th-term test is stated in one direction only and can never certify convergence.
"The terms go to zero, so the sum is zero." The terms going to zero says the later contributions are small. The total is still the total of everything, and $\sum \tfrac{1}{2^n}$ has terms going to zero and sum $1$.
Using the first term of the sequence instead of the first term of the series. $\sum_{n=3}^{\infty} 2^{-n}$ has $a = \tfrac18$, not $\tfrac12$. Write out the first term the series actually contains before touching the formula.
Applying $\dfrac{a}{1 - r}$ when $|r| \ge 1$. With $r = 2$ and $a = 1$ it returns $-1$: a negative total for a sum of positive terms. That absurd answer is the formula doing exactly what it was told outside its hypothesis, and it is worth meeting once so it is recognisable later.
$\displaystyle\sum_{n=2}^{\infty} \frac{3}{4^n}$: the ratio is $\tfrac14$, so it is geometric and $|r| < 1$.
Check the hypothesis first.
The first term actually present is the one at $n = 2$: $\tfrac{3}{16}$.
Not $\tfrac34$.
So the sum is $\dfrac{3/16}{1 - 1/4} = \dfrac{3/16}{3/4} = \tfrac14$.
$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ splits as $\dfrac1n - \dfrac{1}{n+1}$.
Partial fractions, from unit 1.
So $s_N = \left(1 - \tfrac12\right) + \left(\tfrac12 - \tfrac13\right) + \cdots = 1 - \dfrac{1}{N+1}$.
Everything in the middle cancels.
As $N$ grows that approaches $1$, so the series converges to $1$ — and here, unusually, we know the sum and not merely that there is one.
Each term is $-\tfrac12$ times the one before, so $r = -\tfrac12$ and $|r| = \tfrac12 < 1$.
A negative ratio is allowed; its size is what matters.
The first term present is $4$, so the sum is $\dfrac{4}{1 - (-1/2)} = \dfrac{4}{3/2}$.
Mind the double negative in the denominator.
$= \tfrac83$. Notice it lies between the first two partial sums, $4$ and $2$, as an alternating series always does.
The series $1 + 2 + 4 + \cdots$ doubles each time. Complete the table of terms and partial sums.
| Term | Partial sum | |
|---|---|---|
| $k = 0$ | ||
| $k = 1$ | ||
| $k = 2$ |
Put the steps of summing $\dfrac{1}{4} + \dfrac{1}{16} + \dfrac{1}{64} + \cdots$ into the order you do them.
Number the steps in order (write the number in the box):
Match each geometric series to its common ratio.
| $r = \tfrac12$ | $r = \tfrac13$ | $r = \tfrac32$ | $r = -\tfrac12$ | |
|---|---|---|---|---|
| $2 + 1 + \tfrac12 + \cdots$ | ||||
| $9 + 3 + 1 + \cdots$ | ||||
| $4 + 6 + 9 + \cdots$ | ||||
| $8 - 4 + 2 - \cdots$ |
The sequence $a_n = \dfrac{6}{n}$ converges to $0$. What follows about the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{6}{n}$?
For $6 + 3 + \dfrac{3}{2} + \dfrac{3}{4} + \cdots$, give the sum of the whole series, then the sum of the same series with its first two terms removed.
whole series: p, after removing two terms: q
Write the repeating decimal $0.444\ldots$ as an exact fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The series $5 + 10 + 20 + \cdots$ doubles each time. Complete the table of terms and partial sums.
| Term | Partial sum | |
|---|---|---|
| $k = 0$ | ||
| $k = 1$ | ||
| $k = 2$ |
You can say what convergence of a series means and sum a geometric one. Without looking: which list of numbers has to settle for a series to converge, and what is $a$ in $\dfrac{a}{1-r}$ when the series starts at $n = 3$?
9. Your turn: $4 - 2 + 1 - \tfrac12 + \cdots$, step 3