AP Calculus BC

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

The sum of 1 over n plus the square root of n diverges. The square root grows more slowly than n, so it does not change the dominant behaviour: the terms still act like 1 over n, and limit comparison with the harmonic series gives a ratio of 1.

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

Diverges

Settled by the limit comparison test.

The lower-order term does not matter

lim⁡n→∞1n+n1n=lim⁡n→∞nn+n=lim⁡n→∞11+1n=1\lim_{n \to \infty}\frac{\frac{1}{n+\sqrt{n}}}{\frac{1}{n}} = \lim_{n \to \infty}\frac{n}{n+\sqrt{n}} = \lim_{n \to \infty}\frac{1}{1+\frac{1}{\sqrt{n}}} = 1

A ratio of 11 ties this series to the harmonic series, which diverges.

Direct comparison is unavailable in the useful direction: these terms are SMALLER than 1n\frac{1}{n}, and being smaller than a divergent series proves nothing. Limit comparison sidesteps that entirely.

The mistakes students make

  • Using direct comparison with 1n\frac{1}{n} to prove divergence. The inequality runs the wrong way.
  • Treating n\sqrt{n} as the dominant term. It grows far more slowly than nn.

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?

No. It behaves like the harmonic series, and limit comparison gives a ratio of 11.

Why not direct comparison?

The terms are smaller than 1n\frac{1}{n}, and being smaller than a divergent series is not evidence either way.