AP Calculus BC

Does the Sum of sqrt(n)/(n^2+1) Converge?

Q: Does the sum of sqrt(n)/(n^2+1) converge?

Yes. The effective p is \frac{3}{2} > 1.

Q: How do I find the effective p?

Subtract the numerator's degree from the denominator's: 2 - \frac{1}{2} = \frac{3}{2}.

The sum of the square root of n over n squared plus 1 converges. Subtracting the degrees, one half on top and 2 on the bottom, leaves an effective p of three halves, which is greater than 1. Limit comparison with 1 over n to the three halves confirms it.

\sum_{n=1}^{\infty}\frac{\sqrt{n}}{n^{2}+1}

Converges

Settled by the limit comparison test.

Subtract the degrees

Treat the square root as the power $\frac{1}{2}$ , then compare the leading behaviour.

\frac{\sqrt{n}}{n^{2}+1} \approx \frac{n^{1/2}}{n^{2}} = \frac{1}{n^{3/2}}

\lim_{n \to \infty}\frac{\frac{\sqrt{n}}{n^{2}+1}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty}\frac{n^{2}}{n^{2}+1} = 1

Since $\sum \frac{1}{n^{3/2}}$ converges as a $p$ -series with $p = \frac{3}{2} > 1$ , so does this one.

The mistakes students make

Reading the square root as degree $2$ rather than $\frac{1}{2}$ .
Comparing with $\frac{1}{n^{2}}$ and ignoring the numerator. The numerator raises the effective exponent from $2$ down to $\frac{3}{2}$ .

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.