$L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$

Then:

• If $L < 1$: series converges absolutely
• If $L > 1$ (or $L = \infty$): series diverges
• If $L = 1$: inconclusive (test tells you nothing!)

💡 Key Idea: Compare each term to the previous one. If ratio approaches something less than 1, terms shrink fast enough!

Why It Works

If $\frac{a_{n+1}}{a_n} \to r < 1$, then eventually:

$a_{n+1} \approx r \cdot a_n$

This behaves like a geometric series with ratio $r$!

Geometric series with $|r| < 1$ converge.

Example 1: Factorial Series

Test $\sum_{n=1}^{\infty} \frac{n!}{3^n}$ for convergence.

Step 1: Set up ratio

$a_n = \frac{n!}{3^n}, \quad a_{n+1} = \frac{(n+1)!}{3^{n+1}}$

Step 2: Compute $\frac{a_{n+1}}{a_n}$

$\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{3^{n+1}}}{\frac{n!}{3^n}}$

$= \frac{(n+1)!}{3^{n+1}} \cdot \frac{3^n}{n!}$

$= \frac{(n+1)!}{n!} \cdot \frac{3^n}{3^{n+1}}$

$= (n+1) \cdot \frac{1}{3} = \frac{n+1}{3}$

Step 3: Take limit

$L = \lim_{n \to \infty} \frac{n+1}{3} = \infty$

Step 4: Conclusion

Since $L = \infty > 1$:

By Ratio Test, the series diverges.

(Factorials grow faster than exponentials!)

Example 2: Exponential Series

Test $\sum_{n=0}^{\infty} \frac{2^n}{n!}$ for convergence.

Step 1: Set up

$a_n = \frac{2^n}{n!}, \quad a_{n+1} = \frac{2^{n+1}}{(n+1)!}$

Step 2: Compute ratio

$\frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}$

$= \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n}$

$= \frac{2^{n+1}}{2^n} \cdot \frac{n!}{(n+1)!}$

$= 2 \cdot \frac{1}{n+1} = \frac{2}{n+1}$

Step 3: Take limit

$L = \lim_{n \to \infty} \frac{2}{n+1} = 0$

Step 4: Conclusion

Since $L = 0 < 1$:

By Ratio Test, the series converges absolutely.

(This is actually the Taylor series for $e^2$!)

Example 3: When Ratio Test Fails

Test $\sum_{n=1}^{\infty} \frac{1}{n^2}$ using Ratio Test.

Compute ratio:

$\frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}} = \frac{n^2}{(n+1)^2}$

$= \left(\frac{n}{n+1}\right)^2 = \left(\frac{1}{1+\frac{1}{n}}\right)^2$

Take limit:

$L = \lim_{n \to \infty} \left(\frac{1}{1+\frac{1}{n}}\right)^2 = 1$

Conclusion: $L = 1$, so Ratio Test is inconclusive!

(But we know from p-series test that this converges.)

⚠️ When $L = 1$, use a different test!

Ratio Test for Series with Powers

For $\sum a_n x^n$ (power series), the Ratio Test finds the radius of convergence!

We'll cover this more in the Power Series topic.

The Root Test

Given $\sum a_n$ with $a_n \geq 0$, compute:

$L = \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} (a_n)^{1/n}$

Then:

• If $L < 1$: series converges absolutely
• If $L > 1$ (or $L = \infty$): series diverges
• If $L = 1$: inconclusive

When to Use Root Test

Best for series with $n$th powers:

• $(\text{something})^n$
• Terms like $\left(\frac{2n+1}{3n-2}\right)^n$

Example 4: Use Root Test

Test $\sum_{n=1}^{\infty} \left(\frac{2n}{3n+1}\right)^n$ for convergence.

Step 1: Compute $\sqrt[n]{a_n}$

$a_n = \left(\frac{2n}{3n+1}\right)^n$

$\sqrt[n]{a_n} = \left[\left(\frac{2n}{3n+1}\right)^n\right]^{1/n} = \frac{2n}{3n+1}$

Step 2: Take limit

$L = \lim_{n \to \infty} \frac{2n}{3n+1} = \lim_{n \to \infty} \frac{2}{3 + \frac{1}{n}} = \frac{2}{3}$

Step 3: Conclusion

Since $L = \frac{2}{3} < 1$:

By Root Test, the series converges absolutely.

Example 5: Root Test with Exponentials

Test $\sum_{n=1}^{\infty} \frac{n^n}{3^{n^2}}$ for convergence.

Step 1: Compute $\sqrt[n]{a_n}$

$a_n = \frac{n^n}{3^{n^2}}$

$\sqrt[n]{a_n} = \left(\frac{n^n}{3^{n^2}}\right)^{1/n} = \frac{n}{3^n}$

Step 2: Take limit

$L = \lim_{n \to \infty} \frac{n}{3^n}$

This is $\frac{\infty}{\infty}$ form. Use L'Hôpital's Rule:

$= \lim_{n \to \infty} \frac{1}{3^n \ln 3} = 0$

(Exponentials grow much faster than polynomials!)

Step 3: Conclusion

Since $L = 0 < 1$:

By Root Test, the series converges absolutely.

Ratio Test vs Root Test

Ratio Test:

• Best for factorials: $n!$, $(2n)!$
• Best for products: $n \cdot 2^n$
• Generally easier to compute

Root Test:

• Best for $n$th powers: $(\text{stuff})^n$
• Best when taking $n$th root simplifies nicely
• Less commonly used

💡 If both apply, use Ratio Test (usually easier)!

Important Limits for These Tests

Useful limits to know:

$\lim_{n \to \infty} \frac{n!}{n^n} = 0$

$\lim_{n \to \infty} \frac{n^k}{a^n} = 0$ (for any $k$ and $a > 1$)

$\lim_{n \to \infty} n^{1/n} = 1$

$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$

Example 6: Ratio Test with Powers

Test $\sum_{n=1}^{\infty} \frac{5^n n^2}{n!}$ for convergence.

Compute ratio:

$\frac{a_{n+1}}{a_n} = \frac{\frac{5^{n+1}(n+1)^2}{(n+1)!}}{\frac{5^n n^2}{n!}}$

$= \frac{5^{n+1}(n+1)^2}{(n+1)!} \cdot \frac{n!}{5^n n^2}$

$= 5 \cdot \frac{(n+1)^2}{n^2} \cdot \frac{n!}{(n+1)!}$

$= 5 \cdot \frac{(n+1)^2}{n^2} \cdot \frac{1}{n+1}$

$= 5 \cdot \frac{n+1}{n^2}$

Wait, let me recalculate:

$= 5 \cdot \left(\frac{n+1}{n}\right)^2 \cdot \frac{1}{n+1} = \frac{5(n+1)}{n^2}$

Hmm, that's not right either. Let me be more careful:

$\frac{(n+1)^2 n!}{n^2 (n+1)!} = \frac{(n+1)^2}{n^2} \cdot \frac{1}{n+1} = \frac{n+1}{n^2}$

So: $\frac{a_{n+1}}{a_n} = 5 \cdot \frac{n+1}{n^2}$

Actually, let me recalculate more carefully:

$\frac{(n+1)!}{n!} = n+1$

So: $\frac{n!}{(n+1)!} = \frac{1}{n+1}$

$\frac{a_{n+1}}{a_n} = 5 \cdot \frac{(n+1)^2}{n^2(n+1)} = 5 \cdot \frac{n+1}{n^2}$

Taking limit: This goes to 0.

Actually, let me redo this completely:

$\frac{a_{n+1}}{a_n} = \frac{5 \cdot (n+1)^2 \cdot n!}{(n+1)! \cdot n^2 \cdot 1} = \frac{5(n+1)^2 n!}{(n+1) \cdot n! \cdot n^2} = \frac{5(n+1)}{n^2}$

$L = \lim_{n \to \infty} \frac{5(n+1)}{n^2} = \lim_{n \to \infty} \frac{5n}{n^2} = \lim_{n \to \infty} \frac{5}{n} = 0$

Since $L = 0 < 1$, the series converges absolutely.

⚠️ Common Mistakes

Mistake 1: Wrong Ratio Direction

WRONG: $\frac{a_n}{a_{n+1}}$ (backwards!)

RIGHT: $\frac{a_{n+1}}{a_n}$ (next term over current term)

Mistake 2: Simplifying Factorials Wrong

$(n+1)! = (n+1) \cdot n!$, NOT $n! + 1$

$\frac{(n+1)!}{n!} = n+1$

Mistake 3: Thinking $L=1$ Means Diverges

When $L = 1$, the test is inconclusive!

Could converge or diverge - need a different test.

Mistake 4: Forgetting Absolute Value for Negative Terms

Ratio and Root Tests test for absolute convergence.

If series has negative terms, these tests tell you about $\sum |a_n|$.

📝 Practice Strategy

1. Look for factorials: $n!$, $(2n)!$ → use Ratio Test
2. Look for $n$th powers: $(\text{stuff})^n$ → consider Root Test
3. Simplify ratio carefully: Cancel common factors, watch factorials
4. Remember: $L < 1$ converges, $L > 1$ diverges, $L = 1$ inconclusive
5. When $L = 1$: Try comparison test, integral test, or p-series
6. For limits: Use L'Hôpital's if needed, remember $\frac{n^k}{a^n} \to 0$
7. Check your algebra: Factorial mistakes are very common!

Step 2: Compute ratio

$\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)! \cdot 3^{n+1}} \cdot \frac{n! \cdot 3^n}{n^n}$

$= \frac{(n+1)^{n+1}}{n^n} \cdot \frac{n!}{(n+1)!} \cdot \frac{3^n}{3^{n+1}}$

$= \frac{(n+1)^{n+1}}{n^n} \cdot \frac{1}{n+1} \cdot \frac{1}{3}$

$= \frac{(n+1)^n}{n^n} \cdot \frac{1}{3}$

$= \frac{1}{3}\left(\frac{n+1}{n}\right)^n = \frac{1}{3}\left(1 + \frac{1}{n}\right)^n$

Step 3: Take limit

$L = \lim_{n \to \infty} \frac{1}{3}\left(1 + \frac{1}{n}\right)^n$

We know: $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$

$L = \frac{e}{3} \approx \frac{2.718}{3} \approx 0.906$

Step 4: Conclusion

Since $L = \frac{e}{3} < 1$:

By the Ratio Test, the series converges absolutely.

$a_n = \left(\frac{n^2}{e^n}\right)^n$

$\sqrt[n]{a_n} = \left[\left(\frac{n^2}{e^n}\right)^n\right]^{1/n} = \frac{n^2}{e^n}$

Step 2: Take limit

$L = \lim_{n \to \infty} \frac{n^2}{e^n}$

This is $\frac{\infty}{\infty}$ form. Use L'Hôpital's Rule (twice):

$= \lim_{n \to \infty} \frac{2n}{e^n}$ (still $\frac{\infty}{\infty}$)

$= \lim_{n \to \infty} \frac{2}{e^n} = 0$

Step 3: Conclusion

Since $L = 0 < 1$:

By the Root Test, the series converges absolutely.

Note: Exponentials always dominate polynomials!