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  1. AP Calculus
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What is the formula for the Ratio Test?

L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|L=limn→∞​∣an​an+1​​∣

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What is the formula for the Ratio Test?

L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|L=limn→∞​∣an​an+1​​∣

If L<1L < 1L<1 in the Ratio Test, what does this imply?

The series converges.

If L>1L > 1L>1 in the Ratio Test, what does this imply?

The series diverges.

If L=1L = 1L=1 in the Ratio Test, what does this imply?

The test is indeterminate.

Steps to apply the Ratio Test to ∑an\sum a_n∑an​?

  1. Find an+1a_{n+1}an+1​. 2. Compute lim⁡n→∞∣an+1an∣=L\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = Llimn→∞​∣an​an+1​​∣=L. 3. If L<1L < 1L<1, converges; L>1L > 1L>1, diverges; L=1L = 1L=1, inconclusive.

How do you find an+1a_{n+1}an+1​ given ana_nan​?

Replace every 'n' in the expression for ana_nan​ with '(n+1)'.

What should you do if the Ratio Test is indeterminate?

Use a different convergence/divergence test (e.g., Integral Test, Comparison Test).

Given ∑n!nn\sum \frac{n!}{n^n}∑nnn!​, find an+1an\frac{a_{n+1}}{a_n}an​an+1​​.

an+1an=(n+1)!(n+1)n+1⋅nnn!=(n+1)nn(n+1)n+1=nn(n+1)n\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{(n+1)n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n}an​an+1​​=(n+1)n+1(n+1)!​⋅n!nn​=(n+1)n+1(n+1)nn​=(n+1)nnn​

Explain the purpose of the Ratio Test.

To determine whether an infinite series converges or diverges by analyzing the ratio of consecutive terms.

What does the Ratio Test tell us when L = 0?

The series converges absolutely.

Why is the absolute value important in the Ratio Test?

It handles series with alternating signs, ensuring the limit reflects the magnitude of term changes.