Handle constants and finite prefixes
Scaling a series and changing its first terms
The faint outlined bars are the terms \(\frac{1}{n^{2}}\), and the faint curve below is their running total, which levels off near 1.6. The teal bars and curve show a second series.
Move the slider \(c\). Every teal term is \(c\) times the faint term, so the teal total is \(c\) times the faint total. At \(c=5\), the teal curve levels off near 8.2, so the factor changes where the curve levels off but not whether it does.
Press the button that replaces the first four terms by 3. The factor returns to 1, and the first four terms now add to 12 instead of about 1.42. After N = 4, both series add the same terms, so the orange gap between the curves stays near 10.6 and the teal curve still levels off.
Only the far end of the series decides
The faint curve is the running total of any series. For the teal curve, the first four terms are replaced by terms that add to s, and every later term is multiplied by c.
If the faint curve levels off at the dashed height \(M\), the teal curve levels off at a second height \(\cb{M'}\). Move the sliders: \(c\) stretches the teal curve after \(N=4\), and \(s\) lifts it, but it still levels off.
Turn on the toggle. The faint curve now rises without bound, and so does the teal curve, because a stretched or lifted copy of it also passes every bound.
Notation
Multiplying every term by \(c\) multiplies every total by \(c\).
\[\sum_{n=1}^{N}c\,a_{n}=c\sum_{n=1}^{N}a_{n}\]Suppose \(c\) is not 0. If the old totals approach a limit \(L\), the new totals approach \(cL\). If the old totals have no limit, neither do the new ones, because you can divide by \(c\).
\[\sum_{n=1}^{\infty}c\,a_{n}=c\sum_{n=1}^{\infty}a_{n}\]Split the series after the first \(m\) terms. The part that follows is called the tail.
\[\begin{aligned}\sum_{n=1}^{\infty}a_{n}&=\underbrace{a_{1}+\cdots+a_{m}}_{\text{one fixed number}}\\ &\quad+\underbrace{\sum_{n=m+1}^{\infty}a_{n}}_{\text{the tail}}\end{aligned}\]Adding a fixed number to the totals does not change whether they approach a limit. The series converges if its tail converges, and it diverges if its tail diverges.
Adding, deleting or changing finitely many terms changes only the fixed number, so the sum changes but convergence does not.
When it fails
Set \(c=0\) for a series that diverges. Every term becomes 0, so every total is 0 and the new series converges. For this reason the rule needs \(c\neq0\).
The rule covers only finitely many changes. If you replace every term \(\frac{1}{n^{2}}\) by \(\frac1n\), a convergent series becomes divergent.
Example 1: A constant and a late start
Decide whether \(\displaystyle\sum_{n=3}^{\infty}\frac{6}{n^{3}}\) converges.
The teal bars are the terms \(\frac{6}{n^{3}}\), and the faint bars are \(\frac{1}{n^{3}}\). Take out the factor 6, because a factor that is not 0 does not change convergence.
\[\sum_{n=3}^{\infty}\frac{6}{n^{3}}=6\sum_{n=3}^{\infty}\frac{1}{n^{3}}\]The sum starts at \(n=3\), so it lacks the first two terms of the p-series, 1 and \(\frac18\). Deleting two terms changes the sum but not convergence.
\[\sum_{n=1}^{\infty}\frac{1}{n^{3}}=1+\frac18+\sum_{n=3}^{\infty}\frac{1}{n^{3}}\]The p-series has \(p=3\), which is greater than 1, so it converges. The given series converges too, and its teal totals level off below the faint ones.
Example 2: A very small constant
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{1000\sqrt{n}}\) converges.
Take out the constant factor.
\[\sum_{n=1}^{\infty}\frac{1}{1000\sqrt{n}}=\frac{1}{1000}\sum_{n=1}^{\infty}\frac{1}{n^{1/2}}\]A tempting wrong move: The factor \(\frac1{1000}\) makes every term tiny, so the total stays small and the series converges.
\[\begin{aligned}&\tfrac{1}{1000}\cdot\text{(large total)}\\ &\Rightarrow\ \text{small total}\end{aligned}\]The two curves have the same shape, because each teal total is one thousandth of a total for \(\frac1{\sqrt n}\), and those have no bound. After 251 000 terms, the upper total passes 1000, so the teal total passes 1.
The p-series has \(p=\frac12\), which is not greater than 1, so it diverges. The factor \(\frac1{1000}\) does not change this, so the given series diverges.
Practice
A series has the terms 500, \(-40\), 1000 and 7 for \(n=1\) to 4. For \(n\ge5\), the terms are \(\dfrac{9}{n^{6/5}}\). Decide whether it converges.
Only the terms for large \(n\) decide. Split off the first four terms.
Solution
The first four terms add to one fixed number.
\[500-40+1000+7=1467\]Each teal bar is 9 times the faint outlined bar \(\frac{1}{n^{6/5}}\), and \(p=\frac65\) is greater than 1, so the tail converges.
\[\sum_{n=5}^{\infty}\frac{9}{n^{6/5}}=9\sum_{n=5}^{\infty}\frac{1}{n^{6/5}}\]A fixed number plus a convergent tail converges, so the series converges.
Questions
The faint curve is the running total of \(\frac{1}{n^{3}}\), and it levels off near 1.2. The orange curve is the running total of a second series.
The orange series has every term multiplied by 5. About where does the orange curve level off?
- Near 6
- Near 1.2
- Near 5
Answer
Near 6
The first three terms of the orange series are replaced by 4, 4 and 4. What does the gap between the two curves do for N ≥ 3?
- It stays the same, about 10.8.
- It grows without bound.
- It shrinks to 0.
Answer
It stays the same, about 10.8.
Which of these series converges?
- \(\displaystyle\sum_{n=50}^{\infty}\frac{10^{8}}{n^{11/10}}\)
- \(\displaystyle\sum_{n=1}^{\infty}\frac{10^{-6}}{n^{3/4}}\)
- \(\displaystyle\sum_{n=2000}^{\infty}\frac{1}{n^{1/2}}\)
- \(\displaystyle\sum_{n=1}^{\infty}\frac{5}{n}\)
Answer
\(\displaystyle\sum_{n=50}^{\infty}\frac{10^{8}}{n^{11/10}}\)
Summary
A constant factor that is not 0 multiplies every total, and changing finitely many terms adds one fixed number to the later totals.
Neither change decides whether the totals approach a limit, because only the terms far out decide that.
When a series has a factor in front or a different start, you take out the factor and compare the rest with a p-series.
This is one of four skills in topic 10.5 of Which & Why: AP Calculus.