AP Calculus BC

Does the Sum of 1/sqrt(n) Converge? No

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

No. It is the p-series with p = \frac{1}{2}, and convergence requires p > 1.

Q: How does it compare with the harmonic series?

Its terms are larger, since \frac{1}{\sqrt{n}} > \frac{1}{n} for n > 1, so direct comparison with the divergent harmonic series also proves divergence.

The sum of 1 over the square root of n diverges. Writing the term as n to the power negative one half shows it is the p-series with p equal to one half, which is less than 1, so it diverges. Its terms shrink to 0, but more slowly than the harmonic series.

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

Diverges

Settled by the p-series test.

Spotting the p

A radical hides the exponent. Rewriting it as a power makes the $p$ -series match visible.

\frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}} = n^{-1/2} \implies p = \frac{1}{2}

Since $\frac{1}{2} \le 1$ , the series diverges. It is in fact "more divergent" than the harmonic series, because its terms are larger: $\frac{1}{\sqrt{n}} > \frac{1}{n}$ for every $n > 1$ .

A useful direct comparison

Because its terms exceed the harmonic terms and the harmonic series already diverges, direct comparison gives the same answer without the p-series rule. Two routes, one conclusion.

The mistakes students make

Reading the exponent as $2$ because a square root is involved. A square root is the power $\frac{1}{2}$ , not $2$ .
Assuming convergence because the terms tend to $0$ . They do, and it still diverges.
Comparing DOWNWARD to conclude divergence. To prove divergence by direct comparison the terms must be at least as large as a known divergent series, which is the case here.

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/sqrt(n) converge?

No. It is the $p$ -series with $p = \frac{1}{2}$ , and convergence requires $p > 1$ .

How does it compare with the harmonic series?

Its terms are larger, since $\frac{1}{\sqrt{n}} > \frac{1}{n}$ for $n > 1$ , so direct comparison with the divergent harmonic series also proves divergence.