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  2. AP Calculus
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Define a p-series.

A series of the form ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^{p}}∑n=1∞​np1​, where p is a constant.

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Define a p-series.

A series of the form ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^{p}}∑n=1∞​np1​, where p is a constant.

Define a harmonic series.

A p-series where p = 1, represented as ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​.

What does it mean for a series to converge?

The sum of the infinite terms approaches a finite value.

What does it mean for a series to diverge?

The sum of the infinite terms does not approach a finite value; it tends to infinity or oscillates.

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​

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.