Calculus II built a whole toolkit for one question. Does an infinite sum settle down to a finite number, or does it run away? The toolkit has sequences, partial sums, the geometric series formula, and the ratio test. None of it gets rebuilt here. It gets reused, almost at once. The very next section points the ratio test at a series whose terms contain a variable instead of a fixed number. That is all the machinery a power series needs to report how far it can be trusted.
A sequence converges when its values keep walking toward one particular number and never wander back off. They might never arrive there exactly.
A series is a running total of a sequence of add-ons. It converges exactly when the add-ons shrink fast enough that the pile-up loses to the shrinking.
The ratio test makes "shrinking fast enough" precise. Divide each term into the next one, and watch where that ratio settles.
Here is a common confusion. Take the sequence (-1)^n. It does not fail to converge because it is random. It fails because it never settles near any single value. It bounces forever between 1 and -1.
A sequence a_n converges to a limit L when its terms get arbitrarily close to L and stay close as n grows without bound.
An infinite series \sum a_n converges when its sequence of partial sums, S_N=\sum_{n=1}^N a_n, itself converges to a finite limit. The value of the series is defined to be that limit.
A geometric series \sum_{n=0}^\infty ar^n converges exactly when |r|<1. In that case its value is \dfrac{a}{1-r}. For any other r, the terms do not shrink fast enough, or do not shrink at all, for the running total to settle.
The ratio test decides convergence for a much wider class of series. It compares consecutive terms directly. Let L=\displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|. If L<1, the series converges absolutely. If L>1, it diverges. If L=1, the test gives no answer at all.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
A sequence whose terms grow without bound as n\to\infty is said to...
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"Converge" is reserved for settling at a finite number -- infinity is not a finite limit.
Oscillation means bouncing between values, not racing off in one direction.
An unbounded sequence cannot be bounded.
Why
A sequence with no finite limit diverges; "diverges to infinity" specifically describes the unbounded-growth case, as opposed to oscillating divergence.
Convergence always means settling at a finite number.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\lim_{n\to\infty} \dfrac{7n+3}{n+9}.
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Why
Divide top and bottom by n: (7+3/n)/(1+9/n) \to (7+0)/(1+0) = 7.
Correct answer
Also accepted:
Why
Divide numerator and denominator by the highest power of n present.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
The first term of a geometric series is a=20 and the common ratio is r=0.75. Find the sum.
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Why
S=20/(1-0.75)=20/0.25=80.
Correct answer
Also accepted:
Why
Substitute directly into a/(1-r).
From Calculus II — Integral Calculus & Infinite Series
Why does r^n \to 0 exactly when |r|<1, the fact the geometric series formula relies on?
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The size of r is exactly what "small" here needs to mean -- it is not automatic.
Comparing n to r directly does not capture why the limit is 0.
r=2 gives r^n\to\infty, a direct counterexample -- it is not true for every r.
Why
Each multiplication by r (with |r|<1) shrinks the magnitude further; repeated forever, the magnitude is squeezed to 0. For |r|\ge1, this shrinking never happens.
Think about what repeatedly multiplying by a fraction less than 1 in size does over and over.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
For a_n = \dfrac{2^n}{n!}, compute \lim_{n\to\infty} |a_{n+1}/a_n|.
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Why
\dfrac{a_{n+1}}{a_n}=\dfrac{2}{n+1}\to 0 as n\to\infty.
Correct answer
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Why
\dfrac{a_{n+1}}{a_n} = \dfrac{2^{n+1}}{(n+1)!}\cdot\dfrac{n!}{2^n} = \dfrac{2}{n+1}
From Calculus II — Integral Calculus & Infinite Series
If \lim_{n\to\infty} |a_{n+1}/a_n| = L with L<1, what does the Ratio Test conclude?
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L<1 is the CONVERGENCE case, not divergence.
L<1 gives a definite conclusion — absolute convergence — not an inconclusive result.
The Ratio Test proves the stronger property (absolute convergence), never merely conditional convergence.
Why
L<1 means the tail is eventually dominated by a convergent geometric series, forcing absolute convergence.
Terms and their running total, side by side
Toggle the partial-sums view and drag the ratio slider past and below 1 -- watch the running total settle down or run away, exactly matching what the ratio test predicts before you ever compute anything.
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Worked example
A sequence limit by dividing through
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1
Find \displaystyle\lim_{n\to\infty}\frac{3n-1}{n+2}. Divide the numerator and denominator by n, the highest power present: \frac{3n-1}{n+2} = \frac{3-\tfrac1n}{1+\tfrac2n}
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2
As n\to\infty, both \tfrac1n and \tfrac2n go to 0, leaving \frac{3-0}{1+0} = 3
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3\boxed{\lim_{n\to\infty}\frac{3n-1}{n+2} = 3}
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Worked example
Confirming convergence, then summing the series
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1
Determine whether \displaystyle\sum_{n=0}^\infty 4\left(-\frac13\right)^n converges, and find its value if so. This has the form ar^n with a=4 and r=-\tfrac13. The ratio test on its terms gives exactly L=\left|-\tfrac13\right|=\tfrac13<1, so it converges absolutely.
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2
The value of such a series is \dfrac{a}{1-r}: \frac{4}{1-(-\tfrac13)} = \frac{4}{\tfrac43}
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3\boxed{\sum_{n=0}^\infty 4\left(-\frac13\right)^n = 3}
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Where this goes next
The tool the next section reuses directly
The ratio test, run against numbers, is the tool the next section runs against a series whose terms contain a variable x. The same limit comparison decides how far from its center that series can be trusted.
Further out in this course, the sequence limit reviewed here comes back almost unchanged. It applies to sequences of complex numbers. Distance in the plane stands in for absolute value on the line.
Check your understanding (counts toward your progress)
This series is geometric: \displaystyle\sum_{n=1}^\infty 2\left(-\frac15\right)^n. What does it sum to?
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Same idea, with a negative ratio this time: \dfrac{-2/5}{1-(-1/5)}=\dfrac{-2/5}{6/5}=-\dfrac13.
Correct answer
Also accepted:
Why
The first term of the SUM (at n=1) is 2\cdot(-1/5)=-2/5; use \dfrac{a}{1-r} with that a.
What does \displaystyle\sum_{n=0}^\infty 5\left(\frac14\right)^n converge to?
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\dfrac{5}{1-1/4}=\dfrac{5}{3/4}=\dfrac{20}{3}.
Correct answer
Also accepted:
Why
Use \dfrac{a}{1-r} with a=5, r=1/4.
Does a_n=(-1)^n converge as n\to\infty?
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The terms never come close to 0 — they sit exactly at 1 or -1 forever, never in between.
For half the terms, a_n=-1 — that alone keeps the sequence from ever settling near 1.
a_n equals 1 every other step, which is exactly what stops it from converging to -1.
Why
A convergent sequence must settle near ONE value. a_n alternates forever between 1 and -1 and never settles at all.
Which fact most directly confirms whether \sum(1/2)^n converges?
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The divergence test can only ever confirm DIVERGENCE (when terms don't go to 0) — it never confirms convergence by itself.
p-series comparison is unnecessary machinery here — a geometric series has its own immediate convergence rule.
The geometric series test is exactly the tool built for series of this exact shape.
Why
\sum ar^n converges precisely when |r|<1; here r=1/2, so it converges.
Practice: Sequences, series, and the ratio test (5 problems · doesn’t affect your progress)
A series converges when its...
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All-positive terms say nothing about convergence — \sum1/n is entirely positive and still diverges.
Alternating sign is neither required nor sufficient on its own for convergence.
Bounded terms alone don't guarantee convergence of the SERIES — the sequence 1,1,1,\dots is bounded but \sum1 diverges.
Why
A series' convergence is DEFINED by whether its partial sums approach a finite limit.
Sequence of partial sums approaches a finite limit
Evaluate \displaystyle\lim_{n\to\infty}\frac{7n-2}{n+4}.
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Why
Dividing by n: \dfrac{7-2/n}{1+4/n}\to7.
Correct answer
Also accepted:
Why
Evaluate the series \displaystyle\sum_{n=0}^\infty 3\left(\frac27\right)^n to find its sum.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\dfrac{3}{1-2/7}=\dfrac{3}{5/7}=\dfrac{21}{5}.
Correct answer
Also accepted:
Why
Use \dfrac{a}{1-r} with a=3, r=2/7.
Apply the ratio test to \displaystyle\sum \frac{2^n}{n!}. Report the limit L.
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Why
\lim_{n\to\infty}\dfrac{2}{n+1}=0.
Correct answer
Also accepted:
Why
\dfrac{2^{n+1}/(n+1)!}{2^n/n!}=\dfrac{2}{n+1}.
A student says \sum 2^n/n! diverges because 2^n grows without bound. What's wrong with this reasoning?
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2^n does grow without bound on its own — the reasoning's flaw is ignoring what's in the DENOMINATOR.
The series actually converges, not diverges — the ratio test's L=0 settles this.
n! overtakes 2^n decisively for large n — they don't grow at comparable rates at all.
Why
Looking only at the numerator ignores something. The n! in the denominator grows even faster. That drives the ratio test's limit to 0 and the series to convergence.
n! eventually grows far faster than 2^n, so the ratio test gives L=0 and the series converges