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An infinite sum is the limit of its finite partial sums; the geometric and telescoping families are the two where that limit is exact.
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 will be able to state the definition of a series as the limit of its partial sums, sum a geometric series after checking $|r| < 1$ and using the first term the series actually has, recognise and sum a telescoping series by writing out enough terms to see what survives at each end, apply the divergence test in the one direction it runs and say why its converse fails, give the harmonic series as the standing counterexample, and say why rearranging or regrouping an infinite sum is not automatically permitted.
Addition takes two numbers and returns one, and by repetition it takes any finite list. There is no operation that takes an infinite list, and nothing in arithmetic says what $1 + \tfrac12 + \tfrac14 + \cdots$ should mean.
So it has to be defined, and the definition borrows the machinery of the last lesson: add the first $N$ terms — an ordinary finite sum — and let $N \to \infty$. The infinity lives entirely in the limit. That is why a series can fail to have a sum, and why manipulating one as though it were a finite sum is unsafe until a theorem says otherwise.
The terms are the numbers $a_n$ being added. The partial sum $s_N = \sum_{n=1}^{N}a_n$ is a finite sum, one for each $N$. The sum of the series is $\lim_{N\to\infty}s_N$ when that exists.
A series is geometric if each term is a fixed multiple $r$ of the one before, and telescoping if each term is a difference $b_k - b_{k+1}$, so that consecutive terms cancel.
Those two are the only families in this course whose partial sums have a closed form. Everything else is settled by a test that decides whether a sum exists without producing it.
Definition. $\displaystyle\sum_{n=1}^{\infty}a_n := \lim_{N\to\infty}s_N$ where $s_N = a_1 + \cdots + a_N$. The series converges if that limit exists and is finite; otherwise it diverges and has no sum.
Geometric series. For $|r| < 1$, $$\sum_{n=0}^{\infty}ar^n = \frac{a}{1 - r},$$ where $a$ is the first term the series actually has. For $|r| \ge 1$ it diverges. The derivation is one line: $s_N - rs_N = a - ar^N$, so $s_N = a\dfrac{1 - r^N}{1-r}$, and $r^N \to 0$ exactly when $|r| < 1$.
Telescoping series. If $a_k = b_k - b_{k+1}$ then $s_N = b_1 - b_{N+1}$, so the series converges exactly when $b_N$ does, to $b_1 - \lim b_N$. Partial fractions is how a term is put into that form.
The divergence test. If $a_n \not\to 0$ then $\sum a_n$ diverges. Proof: $a_N = s_N - s_{N-1}$, and if $s_N \to S$ then $a_N \to S - S = 0$.
And its converse is false. $\sum \dfrac1n$ has terms tending to zero and diverges. Terms going to zero is necessary and nowhere near sufficient. The harmonic series has terms going to zero and diverges; that single example is why the divergence test is stated in one direction only and can never certify convergence.
Linearity. Convergent series may be added and scaled termwise: $\sum(a_n + cb_n) = \sum a_n + c\sum b_n$. Rearranging and regrouping are not covered by this and are dangerous for series that are not absolutely convergent — lesson 23.
Another way: picture
A number line with the partial sums of $\tfrac12 + \tfrac14 + \tfrac18 + \cdots$ marked: $0.5$, $0.75$, $0.875$, $0.9375$. Each step covers half the remaining distance to $1$, so the marks pile up at $1$ without ever reaching it. The series 'equals $1$' means exactly that the marks have $1$ as their limit — a statement about the sequence of marks, not about the last one, of which there is none.
Another way: steps
The terms $\dfrac1n$ go to zero, and the partial sums grow without bound. Both are true, and the second is worth seeing proved, because it is the counterexample the whole unit leans on.
Group the terms in blocks of doubling length: $$1 + \underbrace{\frac12}_{\ge 1/2} + \underbrace{\frac13 + \frac14}_{\ge 1/2} + \underbrace{\frac15 + \cdots + \frac18}_{\ge 1/2} + \cdots$$ Each block has twice as many terms as the last, and each term in it is at least the last term of the block, so each block totals at least $\tfrac12$. There are infinitely many blocks, so the partial sums exceed any bound.
They do so extraordinarily slowly: $s_N \approx \ln N$, so reaching $20$ takes about $2.7 \times 10^{8}$ terms and reaching $100$ takes more terms than there are atoms in a galaxy. No amount of numerical evidence would ever suggest divergence. That is the case for proving things.
And it is why "the terms go to zero" can never be a test for convergence. The terms of a convergent series must go to zero, but so do the terms of this one. What matters is how fast, and that is what every test in the next five lessons measures.
Reading the divergence test backwards. Terms to zero proves nothing. The test rules series out and never in.
Using the wrong first term in a geometric sum. $\sum_{n=2}^{\infty}ar^n$ has first term $ar^2$, not $a$. The formula is (first term)$/(1-r)$, and the first term is whichever one the sum actually starts at.
Using the geometric formula without checking $|r| < 1$. It produces a number for $|r| > 1$, and that number can be negative for a series of positive terms — a hypothesis violated in the open.
Confusing the sequence with the series. $a_n \to 0$ is about the terms; $\sum a_n$ is about the partial sums. Keeping the two objects apart is most of the battle.
Rearranging or regrouping freely. Finite sums allow it and infinite ones may not. $1 - 1 + 1 - 1 + \cdots$ groups to $0$ or to $1$ depending on where the brackets go, and in fact diverges.
Reporting a divergent series' sum as $\infty$. It has no sum. "Diverges to $+\infty$" describes; "equals $\infty$" asserts.
The notation $\sum_{n=1}^{\infty}$ looks like an instruction to add, with a slightly unusual upper limit, and that reading is behind almost every error in this unit. It suggests that the sum is what you would get at the end of the additions, and that the ordinary rules of addition apply.
Neither is so. There is no end of the additions; the sum is the limit of the finite totals, which is a statement about a sequence. And the rules of finite addition do not automatically carry over: commutativity fails for conditionally convergent series, associativity fails for divergent ones. $1 - 1 + 1 - 1 + \cdots$ brackets to $(1-1) + (1-1) + \cdots = 0$ or to $1 + (-1+1) + \cdots = 1$, and it has no sum at all.
This is also the honest answer to the perennial objection about $0.\overline9 = 1$. Nobody is claiming that adding nines forever eventually arrives at $1$. The claim is that the sequence $0.9, 0.99, 0.999, \ldots$ has limit $1$, and that the notation $0.\overline9$ is defined to mean that limit. Once the definition is in view the statement is not surprising and not a trick; it is an arithmetic fact about a sequence.
So read every $\sum_{n=1}^{\infty}$ as "the limit of the partial sums of". The phrase is longer and it keeps the object and its definition attached.
$\displaystyle\sum_{k=1}^{\infty}\frac{1}{k(k+2)}$: decompose, $\dfrac{1}{k(k+2)} = \dfrac12\left(\dfrac1k - \dfrac{1}{k+2}\right)$.
Partial fractions is what makes a term telescoping.
The gap is two, so terms cancel two apart and two survive at each end: $s_N = \dfrac12\left(1 + \dfrac12 - \dfrac{1}{N+1} - \dfrac{1}{N+2}\right)$.
Write out five terms to see which survive.
Letting $N \to \infty$: the sum is $\dfrac12 \cdot \dfrac32 = \dfrac34$. Assuming only one term survives at each end is the usual slip, and writing out the first few terms catches it.
$\displaystyle\sum_{n=3}^{\infty}\frac{2^{n}}{3^{n}}$: the ratio is $\tfrac23$, so it converges.
Check the ratio first.
The first term is the one at $n = 3$: $\left(\tfrac23\right)^3 = \tfrac{8}{27}$.
Not $1$, and not $\tfrac23$.
Sum $= \dfrac{8/27}{1 - 2/3} = \dfrac{8/27}{1/3} = \dfrac{8}{9}$. Starting at $n = 0$ instead would give $3$, and the difference $3 - \tfrac89 = \tfrac{19}{9}$ is exactly the three terms that were skipped.
Both pieces are geometric with ratio below one, so both converge and linearity permits splitting the sum.
Check convergence before splitting, not after.
First piece: first term $\tfrac12$, ratio $\tfrac12$, sum $\dfrac{1/2}{1/2} = 1$.
First term is the $n = 1$ term.
Second: first term $\tfrac13$, ratio $\tfrac13$, sum $\dfrac{1/3}{2/3} = \tfrac12$. Total $\tfrac32$.
Evaluate $\displaystyle\sum_{n=0}^{\infty} 5\left(\frac{1}{4}\right)^{n}$.
Answer:
For $\displaystyle\sum_{k=0}^{\infty}1\cdot 2^{k}$, plot the partial sums $s_1$, $s_2$ and $s_3$ against the number of terms.
Plot your answer on the grid:
The terms of $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{1/6}}$ tend to zero. What does that establish?
Put the steps of the definition of $\displaystyle\sum_{n=1}^{\infty}a_n$ into order.
Number the steps in order (write the number in the box):
For $\displaystyle\sum_{k=1}^{N}\left(\frac{1}{k} - \frac{1}{k+1}\right)$, fill in the partial sums.
| Value | |
|---|---|
| $s_1$ | |
| $s_2$ | |
| $s_{3}$ |
Write the repeating decimal $0.\overline{1} = 0.111\ldots$ as a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Evaluate $\displaystyle\sum_{n=0}^{\infty} 2\left(\frac{2}{3}\right)^{n}$.
Answer:
You can sum the two families that can be summed, and say what a sum is when it cannot. Say in your own words why $0.\overline{9} = 1$ is a statement about a limit rather than about adding nines. Next: the tests that decide convergence without producing the sum.
10. Your turn: $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{2^{n}} + \frac{1}{3^{n}}\right)$, step 3