Order of Convergence - GRE Quantitative Reasoning

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Question

For which values of p is

$\sum_{n=1}^\infty \frac{1}{n^p}$

convergent?

Answer

We can solve this problem quite simply with the integral test. We know that if

$\int_1^{\infty} \frac{1}{x^p} dx$

converges, then our series converges.

We can rewrite the integral as

$\int x^{-p} dx$

and then use our formula for the antiderivative of power functions to get that the integral equals

$\frac{x^{1-p}}{1-p} \Big|_{x=1}^\infty$ .

We know that this only goes to zero if . Subtracting p from both sides, we get

.

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Question

For which values of p is

$\sum_{n=1}^\infty \frac{1}{n^p}$

convergent?

Answer

We can solve this problem quite simply with the integral test. We know that if

$\int_1^{\infty} \frac{1}{x^p} dx$

converges, then our series converges.

We can rewrite the integral as

$\int x^{-p} dx$

and then use our formula for the antiderivative of power functions to get that the integral equals

$\frac{x^{1-p}}{1-p} \Big|_{x=1}^\infty$ .

We know that this only goes to zero if . Subtracting p from both sides, we get

.

Compare your answer with the correct one above

Question

For which values of p is

$\sum_{n=1}^\infty \frac{1}{n^p}$

convergent?

Answer

We can solve this problem quite simply with the integral test. We know that if

$\int_1^{\infty} \frac{1}{x^p} dx$

converges, then our series converges.

We can rewrite the integral as

$\int x^{-p} dx$

and then use our formula for the antiderivative of power functions to get that the integral equals

$\frac{x^{1-p}}{1-p} \Big|_{x=1}^\infty$ .

We know that this only goes to zero if . Subtracting p from both sides, we get

.

Compare your answer with the correct one above

Question

For which values of p is

$\sum_{n=1}^\infty \frac{1}{n^p}$

convergent?

Answer

We can solve this problem quite simply with the integral test. We know that if

$\int_1^{\infty} \frac{1}{x^p} dx$

converges, then our series converges.

We can rewrite the integral as

$\int x^{-p} dx$

and then use our formula for the antiderivative of power functions to get that the integral equals

$\frac{x^{1-p}}{1-p} \Big|_{x=1}^\infty$ .

We know that this only goes to zero if . Subtracting p from both sides, we get

.

Compare your answer with the correct one above

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