AP Calculus BC

Does the Sum of (-1)^n/sqrt(n) Converge?

The sum of negative 1 to the n over the square root of n converges CONDITIONALLY. The magnitudes decrease to 0 so the alternating series test applies, but taking absolute values gives the p-series with p equal to one half, which diverges.

∑n=1∞(−1)nn\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}

Converges

Settled by the alternating series test, and only conditionally.

Both conditions hold

With bn=1nb_{n} = \frac{1}{\sqrt{n}}, the magnitudes are decreasing and tend to 00, so the alternating series test gives convergence.

1n+1<1nandlim⁡n→∞1n=0\frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} \quad \text{and} \quad \lim_{n \to \infty}\frac{1}{\sqrt{n}} = 0

Decay can be slow and still qualify

At n = 1,000,000 the magnitude is still 0.001. The test asks only that the magnitudes decrease to 0, never how fast, which is why slow decay is no obstacle.

The absolute series diverges

Absolute values give ∑1n\sum \frac{1}{\sqrt{n}}, the pp-series with p=12≤1p = \frac{1}{2} \le 1, which diverges. So the convergence is conditional, not absolute.

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

The mistakes students make

  • Concluding divergence because ∑1n\sum \frac{1}{\sqrt{n}} diverges. The absolute series failing rules out ABSOLUTE convergence, not convergence.
  • Reading pp as 22 from the square root. It is 12\frac{1}{2}.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the sum of (-1)^n/sqrt(n) converge?

Yes, conditionally, by the alternating series test.

Why is it not absolutely convergent?

The absolute series is the pp-series with p=12p = \frac{1}{2}, which diverges.