Ratio and Root Tests

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Use exponential and factorial structure to test absolute convergence without overinterpreting the boundary case.

Compare the tail with geometric decay

For a series with eventually nonzero terms, suppose $L=\lim|a_{n+1}/a_n|$ exists. The ratio test gives absolute convergence if $L<1$ and divergence if $L>1$ (including infinity). At $L=1$ , it gives no conclusion.

The root test uses $L=\lim\sqrt[n]{|a_n|}$ when that limit exists, with the same outcomes. A limsup version covers some cases without an ordinary limit, but the simple limit form handles our examples. The absolute values make both tests apply to signed terms.

Why does $L<1$ work? Choose a number $q$ between $L$ and $1$ . Far enough into the sequence, successive sizes shrink at least as fast as a geometric sequence with ratio $q$ . A finite prefix cannot spoil convergence.

Visual guide

VISUAL GUIDEA limiting ratio below one forces geometric decay

For aₙ = 2ⁿ/n!, the next-term ratio is 2/(n + 1), which tends to zero. Early terms need not decrease, but eventually every step shrinks by a uniform factor below one, proving absolute convergence.

Worked example: a factorial denominator

For $a_n=3^n/n!$ ,

\left|\frac{a_{n+1}}{a_n}\right|=\frac3{n+1}\longrightarrow0.

Thus $\sum3^n/n!$ converges absolutely. Cancel factorials before taking limits: $(n+1)!=(n+1)n!$ . Treating a factorial as merely a polynomial would give the wrong growth comparison.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

The ratio-test limit is exactly 1. What can you conclude from that test alone?

The series diverges.

It is inconclusive.

The series converges.

Hint 1 · Find a starting point

The strict cases are L<1 and L>1.

Hint 2 · Take the next step

Both Σ1/n and Σ1/n² have ratio limit 1.

Show the reasoning

Answer: It is inconclusive.

One of those series diverges and the other converges, so another test is needed.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: powers indexed by n

For $a_n=((2n+1)/(3n+2))^n$ , the $n$ th root is $(2n+1)/(3n+2)\to2/3$ . The series converges. The root test removes the outer exponent immediately, avoiding a cumbersome ratio involving both $n$ and $n+1$ .

Both $\sum1/n$ and $\sum1/n^2$ have ratio limit $1$ , yet one diverges and the other converges. The boundary case is genuinely undecided. Switch to comparison or an integral test instead of trying to extract a verdict from $L=1$ .

Practice

Test $\sum n/2^n$ .
Test $\sum n!$ .
What does the root test say about $\sum1/n^3$ ?

Show worked solutions

Ratio $(n+1)/(2n)\to1/2$ , so it converges absolutely.
Ratio $n+1\to\infty$ , so it diverges; its terms also fail to approach zero.
The root tends to $1$ , so this test is inconclusive. The $p$ -series test gives convergence.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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