1. Why we need new tests

All the tests you learned earlier (Integral, Comparison, ‑p‑series, etc.) assume every term is positive. But many useful series zig‑zag above and below zero, for example

$$1-\tfrac12+\tfrac13-\tfrac14+\dots$$

So we need tests that handle sign changes.

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2. Alternating Series & the Alternating Series Test (AST)

Why it works (picture) – partial sums bounce left‑right but the “zig‑zag” steps keep shrinking, so they squeeze toward a single number.

Example A (Alternating harmonic series)

$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac1n$$

$b_n=\tfrac1n$ is decreasing and →0, so it converges by AST.

Example B (Fails AST)

$$\sum_{n=1}^{\infty}(-1)^{n}\frac{1-3n}{n+7}$$

Limit of $b_n$ is 3 ≠ 0, so AST cannot help; the series actually diverges by the basic “limit of a_n” test (often called Test for Divergence).

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3. Absolute vs Conditional Convergence

Big fact: Absolute $\implies$ Convergent (proof uses Comparison Test with $|a_n|$).

• Alternating harmonic series is conditionally convergent – its absolute version is the divergent harmonic series.
• $\displaystyle\sum_{n=1}^{\infty}\frac{\cos n}{n^{2}}$ is absolutely convergent because $|\cos n|\le1$ and $\sum 1/n^{2}$ converges (‑p‑series with $p>1$).

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4. Ratio Test (great when factorials or exponentials appear)

Compute $L=\displaystyle\lim_{n\to\infty}\Bigl|\frac{a_{n+1}}{a_n}\Bigr|$.

Example C 

$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{2^n}$$ $$\left|\frac{a_{n+1}}{a_n}\right| =\frac{(n+1)^2}{2^{\,n+1}}\cdot\frac{2^{\,n}}{n^{2}} \to\frac12<1$$

Series converges absolutely. (Your handwritten page marks it “AC” ✔.)

Example D 

$$\sum_{n=1}^{\infty}\frac{n^n}{n!}$$ $$\frac{a_{n+1}}{a_n}\to e>1$$

Series diverges. (Matches your slide with result “D” for divergent.)

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5. Root Test (handy when the whole term is raised to the $n$-th power)

Compute $L=\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_n|}$.

Same conclusions as the Ratio Test: $L<1$ $\implies$ converge, $L>1$ $\implies$ diverge, $L=1$ $\implies$ inconclusive.

Example E 

$$\sum_{n=1}^{\infty}\Bigl(\frac{2n-3}{3n-2}\Bigr)^{n}$$ $$\sqrt[n]{|a_n|}\;=\;\frac{2n-3}{3n-2}\;\longrightarrow\;\frac23<1$$

So the series converges.

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6. Quick Strategy Cheat‑Sheet

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7. Practice – Try these quick checks

1. $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+2)}$
2. $\displaystyle \sum_{n=1}^{\infty}\frac{(2n)!}{4^{\,n}(n!)^{2}}$
3. $\displaystyle \sum_{n=1}^{\infty}\frac{\sin n}{n^{3/2}}$

(Hint: 1 → AST then absolute test, 2 → Ratio, 3 → compare |sin n| to 1 and use p‑series.)

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8. Key Take‑aways

• Alternating signs + shrinking size $\implies$ convergence (AST).
• Absolute convergence is stronger; if it happens, you’re done.
• Ratio & Root tests are powerhouse tools whenever factorials or nth‑powers show up.
• Always check the limit of $a_n$ first: if it isn’t zero, the series diverges instantly.

Keep this roadmap handy and series tests will feel a lot less mysterious!