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  2. AP Calculus
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Explain the convergence/divergence of a p-series.

A p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^{p}}∑n=1∞​np1​ converges if p > 1 and diverges if p ≤ 1.

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Explain the convergence/divergence of a p-series.

A p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^{p}}∑n=1∞​np1​ converges if p > 1 and diverges if p ≤ 1.

Explain the behavior of the harmonic series.

The harmonic series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​ diverges, even though its terms approach zero.

Why is simplifying the terms important before applying the p-series test?

Simplification ensures the series is in the standard 1np\frac{1}{n^p}np1​ form, allowing for correct identification of 'p' and accurate convergence/divergence determination.

How do you determine convergence/divergence of ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^{2}}∑n=1∞​n21​?

Identify p = 2. Since 2 > 1, the series converges.

How do you determine convergence/divergence of ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}∑n=1∞​n​1​?

Rewrite as ∑n=1∞1n1/2\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}∑n=1∞​n1/21​. Identify p = 1/2. Since 1/2 < 1, the series diverges.

How do you determine convergence/divergence of ∑n=1∞n2n3\sum_{n=1}^{\infty} \frac{n^2}{n^3}∑n=1∞​n3n2​?

Simplify to ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​. Identify p = 1. Since p=1, the series diverges (Harmonic Series).

Determine convergence/divergence of ∑n=1∞1n0.75\sum_{n=1}^{\infty} \frac{1}{n^{0.75}}∑n=1∞​n0.751​

Identify p = 0.75. Since 0.75 < 1, the series diverges.

Determine convergence/divergence of ∑n=1∞n−4\sum_{n=1}^{\infty} n^{-4}∑n=1∞​n−4

Rewrite as ∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{n^4}∑n=1∞​n41​. Identify p = 4. Since 4 > 1, the series converges.

Determine convergence/divergence of ∑n=1∞nn5\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^5}}∑n=1∞​n5​n​

Simplify to ∑n=1∞1n3/2\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}∑n=1∞​n3/21​. Identify p = 3/2. Since 3/2 > 1, the series converges.

Determine convergence/divergence of ∑n=1∞(1n)3\sum_{n=1}^{\infty} (\frac{1}{n})^3∑n=1∞​(n1​)3

Rewrite as ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3}∑n=1∞​n31​. Identify p = 3. Since 3 > 1, the series converges.

What is the general form of a p-series?

∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^{p}}∑n=1∞​np1​

What is the formula for the harmonic series?

∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​