Unit 4 · Lesson 5 of 9

A factorial can outrun an exponential.

Terms involving factorials can initially grow, yet eventually shrink rapidly. Inspect how one term compares with the next.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its successive absolute-term ratio.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase term index n? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Read the model

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

RatiosThreshold 1 · dashed

Term index n5 Parameter a2

successive absolute-term ratio

\left|\frac{u_{6}}{u_5}\right|=\frac{2}{5+1}=0.333333

The test uses a limit, not an early finite ratio. Initial terms may grow before factorial growth eventually dominates.

Assumptions, units, and limits

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

Form the absolute successive-term ratio and take its limit. A limit below one proves absolute convergence; above one or infinity proves divergence. Limit one is inconclusive and requires a different test.

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Keep the reasoning visible

L=\lim\left|\frac{a_{n+1}}{a_n}\right|

The test uses a limit, not an early finite ratio. Initial terms may grow before factorial growth eventually dominates.

What makes this model useful?

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

An early ratio above one does not contradict eventual convergence. The test depends on its limiting value.

A worked example

Calculate the starting scenario

Use term index n = 5 ; parameter a = 2 .
$\left|\frac{u_{6}}{u_5}\right|=\frac{2}{5+1}=0.333333$
The test uses a limit, not an early finite ratio. Initial terms may grow before factorial growth eventually dominates. The checked quantity is successive absolute-term ratio.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the successive absolute-term ratio using term index n 5 , parameter a 2 . State the meaning of its sign and units.

Checked quantity: successive absolute-term ratio · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Read the model

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

RatiosThreshold 1 · dashed

Term index n5 Parameter a2

successive absolute-term ratio

\left|\frac{u_{6}}{u_5}\right|=\frac{2}{5+1}=0.333333

The test uses a limit, not an early finite ratio. Initial terms may grow before factorial growth eventually dominates.

Assumptions, units, and limits

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Calculate and explain

Apply the idea to a new scenario

For parameter $a=3$ at $n=8,$ find the ratio of consecutive terms aⁿ/n!.

Checked quantity: successive absolute-term ratio · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Read the model

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

RatiosThreshold 1 · dashed

Term index n8 Parameter a3

successive absolute-term ratio

\left|\frac{u_{9}}{u_8}\right|=\frac{3}{8+1}=0.333333

The test uses a limit, not an early finite ratio. Initial terms may grow before factorial growth eventually dominates.

Assumptions, units, and limits

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

Read the model

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

RatiosThreshold 1 · dashed

Term index n5 Parameter a2

successive absolute-term ratio

\left|\frac{u_{6}}{u_5}\right|=\frac{2}{5+1}=0.333333

The test uses a limit, not an early finite ratio. Initial terms may grow before factorial growth eventually dominates.

Assumptions, units, and limits

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Form the absolute successive-term ratio and take its limit. A limit below one proves absolute convergence; above one or infinity proves divergence. Limit one is inconclusive and requires a different test.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.