AP Calculus BC

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

The sum of 1 over the square root of n plus 1 diverges. Limit comparison against 1 over the square root of n, the p-series with p equal to one half, gives a ratio of 1, and a p-series with p at most 1 diverges.

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

Diverges

Settled by the limit comparison test.

The obvious comparison points the wrong way

Since n+1>n\sqrt{n}+1 > \sqrt{n}, a larger denominator makes a smaller fraction.

1n+1<1n\frac{1}{\sqrt{n}+1} < \frac{1}{\sqrt{n}}

Direct comparison proves divergence only when the terms are at least as large as those of a known divergent series. Sitting below a divergent series proves nothing, since a series can lie under a divergent one and still converge. This is exactly the situation the limit comparison test was built for.

Limit comparison with the p-series

Take bn=1nb_{n} = \frac{1}{\sqrt{n}}, the dominant behaviour of the denominator, and form the ratio.

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

The limit is finite and positive, so the two series share a verdict. And ∑1n\sum \frac{1}{\sqrt{n}} is the p-series with p=12p = \frac{1}{2}, which is not greater than 11, so it diverges. This series diverges with it.

The number 1 is not what matters

Any limit strictly between zero and infinity gives the same conclusion. Getting exactly 1 is a sign the comparison series was well chosen, nothing more. What the test needs is that the limit exists and is not zero.

The mistakes students make

The first of these is the reason this series appears on exams at all.

  • Using direct comparison with 1n\frac{1}{\sqrt{n}} to claim divergence. The inequality runs the wrong way, because these terms are smaller.
  • Concluding convergence because the terms tend to 00. The nth term test can only prove divergence, never convergence.
  • Treating p=12p = \frac{1}{2} as a convergent case. A p-series converges only when p>1p > 1, and square roots put you below that line.

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)+1) converge?

No, it diverges, by limit comparison with ∑1n\sum \frac{1}{\sqrt{n}}.

Why does direct comparison with 1/sqrt(n) not work here?

Because 1n+1<1n\frac{1}{\sqrt{n}+1} < \frac{1}{\sqrt{n}}. To prove divergence you need terms at or above a divergent series, and these sit below it. A different comparison can still work, since 1n+1≥1n\frac{1}{\sqrt{n}+1} \ge \frac{1}{n} for n≥3n \ge 3, but limit comparison is the cleaner route because it needs no inequality at all.

What does the plus 1 in the denominator change?

Almost nothing. For large nn the n\sqrt{n} dominates, and the ratio of the two terms tends to 11.