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Partial sums, the geometric and $p$-series families, and the term, comparison, ratio and root tests.
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 treat a series as the sequence of its partial sums, sum a geometric series, and settle a series of non-negative terms with the test its shape calls for — the $n$-th term test, comparison with a geometric or $p$-series, the ratio test or the root test — checking each test's hypothesis before its verdict. You will also be able to say why terms tending to zero never proves convergence.
Everything in this lesson is a theorem about sequences in disguise. A series is the sequence of its partial sums, and the tests are monotone convergence and the Cauchy criterion applied to that sequence. Nothing new about infinity is introduced, and nothing needs to be.
Given terms $a_1, a_2, \ldots$, the partial sums are $s_n = a_1 + \cdots + a_n$. The series $\sum a_n$ converges when the sequence $(s_n)$ converges, and its sum is that limit. So a series is not an addition of infinitely many numbers; it is a limit of finitely many additions, and every question about it is a question about $(s_n)$.
The $n$-th term test. If $\sum a_n$ converges then $a_n \to 0$, because $a_n = s_n - s_{n-1} \to S - S = 0$. Used in the contrapositive: terms that do not vanish force divergence. It never proves convergence.
Comparison (for $a_n, b_n \ge 0$). If $a_n \le b_n$ for large $n$: $\sum b_n$ convergent gives $\sum a_n$ convergent, and $\sum a_n$ divergent gives $\sum b_n$ divergent. The proof is monotone convergence — with non-negative terms the partial sums increase, so bounded is the same as convergent.
The ratio test. If $|a_{n+1}/a_n| \to L$: convergence (absolute) when $L < 1$, divergence when $L > 1$, nothing when $L = 1$.
The root test. If $|a_n|^{1/n} \to L$: the same three verdicts. It is strictly stronger than the ratio test, and it is the one the radius of a power series is defined by.
The two comparison families worth memorising.
The harmonic series $\sum 1/n$ is the boundary case, and it diverges: grouping terms into blocks of length $2^k$ gives at least $1/2$ per block, without limit.
Another way: picture
Draw the partial sums as heights. A series with non-negative terms gives a staircase that only climbs; it converges exactly when the staircase has a ceiling. Nothing about infinitely many additions is involved — only whether a climbing sequence is bounded.
Another way: steps
To settle a series:
Comparison is monotone convergence. With $a_n \ge 0$ the partial sums increase; if they are bounded above — which is what domination by a convergent series gives — they converge.
The Cauchy criterion for series reads: $\sum a_n$ converges exactly when for every $\varepsilon$ there is $N$ with $|a_{n+1} + \cdots + a_m| < \varepsilon$ for all $m > n \ge N$. It is the Cauchy criterion for $(s_n)$ written out, and it is what proves the $n$-th term test (take $m = n+1$) and the comparison test at a stroke.
The ratio and root tests are comparisons with a geometric series in disguise. If $|a_n|^{1/n} \to L < 1$, choose $r$ between $L$ and $1$; then $|a_n| \le r^n$ for large $n$, and $\sum r^n$ converges.
Seeing them this way is worth the effort: it says which test can be trusted where, and it explains why $L = 1$ is inconclusive — the comparison being made is with $\sum 1^n$, which says nothing.
| Term contains | Reach for | Because |
|---|---|---|
| $n!$ or a product of $n$ factors | the ratio test | consecutive terms cancel |
| something to the power $n$ | the root test | the root undoes the power |
| a quotient of polynomials | comparison with $1/n^p$ | only the leading powers matter |
| a term not tending to $0$ | the $n$-th term test | divergence, immediately |
| alternating signs | the next lesson | these tests want non-negative terms |
The last row matters. Comparison assumes non-negative terms outright, and the ratio and root tests are really tests for absolute convergence. A series whose signs alternate needs the treatment of the next lesson.
Reading the term test forwards. Terms tending to $0$ does not give convergence. $\sum 1/n$ is the standing counterexample and should be the first thing that comes to mind.
Comparing in the wrong direction. Being smaller than a divergent series proves nothing, and being larger than a convergent one proves nothing either. Only two of the four combinations say anything.
Treating an inconclusive test as a verdict. Ratio limit $1$ means the test is silent, not that the series diverges. Both $\sum 1/n$ and $\sum 1/n^2$ have ratio limit $1$ and different verdicts.
Manipulating a series before it is known to converge. Splitting, reordering or factorising a divergent series produces true-looking lines with no content. The convergence is established first.
For $\sum \dfrac{n!}{n^n}$, the ratio is $\dfrac{(n+1)!}{(n+1)^{n+1}} \cdot \dfrac{n^n}{n!} = \left(\dfrac{n}{n+1}\right)^n$.
The factorial cancels, which is why this test was chosen.
That is $(1 + 1/n)^{-n} \to 1/e$, which is below $1$.
The limit of the ratio, computed.
So the series converges. The value of the sum is not produced and is not needed; convergence is the question.
A verdict, not a value.
For $\sum \dfrac{1}{n^2 + n}$, the terms are positive, so comparison is available.
Check the hypothesis first.
And $\dfrac{1}{n^2 + n} \le \dfrac{1}{n^2}$ for every $n \ge 1$, while $\sum 1/n^2$ converges.
Dominated by a convergent series.
So the series converges. Had the inequality gone the other way it would have said nothing, which is why the direction is checked before the verdict is written.
The direction is the argument.
The terms are positive and contain an $n$-th power, so the root test is the natural choice.
Shape chooses the test.
$\left(\dfrac{2^n}{n^3}\right)^{1/n} = \dfrac{2}{n^{3/n}} \to 2$, since $n^{1/n} \to 1$.
Compute the limit of the root.
The limit exceeds $1$, so the series diverges. The $n$-th term test would have settled it too, since the terms grow without bound — and noticing that first would have saved the computation, which is why step 1 of the recipe is always the term test.
For the series whose $n$-th term is $\dfrac{6}{2^n}$, starting at $n = 1$, give the first three partial sums.
| partial sum | |
|---|---|
| first partial sum | |
| second partial sum | |
| third partial sum |
What is $\displaystyle\sum_{n=1}^{\infty} \dfrac{2}{9^{\,n}}$?
Answer:
Match each series to the test its terms are shaped for.
| the ratio test | the root test | the $n$-th term test | comparison with a $p$-series | |
|---|---|---|---|---|
| $\sum 1/n!$ | ||||
| $\sum 2^n/n^2$ | ||||
| $\sum n/(n+1)$ | ||||
| $\sum 1/n^{3}$ |
Put the steps of a comparison argument — say for $\sum \dfrac{1}{n^{3} + n}$ — into order.
Number the steps in order (write the number in the box):
Does $\sum 1/2^n$ converge?
Here is an argument that $\sum \dfrac{4}{n}$ converges. Mark the one step that is not a theorem.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For the series whose $n$-th term is $\dfrac{5}{2^n}$, starting at $n = 1$, give the first three partial sums.
| partial sum | |
|---|---|
| first partial sum | |
| second partial sum | |
| third partial sum |
You can compute partial sums, sum a geometric series, and choose and apply the test a series calls for. Say in your own words why the harmonic series diverges although its terms tend to zero. Next: what changes when the terms are allowed to change sign.
10. Your turn: settle $\sum \dfrac{2^n}{n^3}$, step 3