When applying a convergence or divergence test, does it matter if $<$ is used instead of $\leq$?

For example, consider the series $\sum_{n=1}^{\infty} \sqrt{n+1} - \sqrt{n}$.

The direct comparison test: If $0 \leq a_n \leq b_n$ for all $n > N \in \mathbb{Z}^+$, then,

$\sum a_n$ converges if $\sum b_n$ converges and $\sum b_n$ diverges if $\sum a_n$ diverges.

Applying it:

Notice that for $n >$ some $N$, $n > \sqrt{n+1} > \sqrt{n+1} - \sqrt{n}$

$\implies 0 < \frac{1}{n} < \frac{1}{\sqrt{n+1} - \sqrt{n}} < \sqrt{n+1} - \sqrt{n}$

We know that $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, so $\sum_{n=1}^{\infty} \sqrt{n+1} - \sqrt{n}$ also diverges.

Is the above correct? Is it correct to apply the test using $<$ instead of $\leq$ even though the direct comparison test uses $\leq$? Or must the inequality not be strict? Does this apply to all the convergence/divergence test that utilize inequalities? Any insight is much appreciated.


1 Answer 1


You are not using $<$ instead of $\leqslant$. You proved that you always have $\frac1n<\sqrt{n+1}-\sqrt n$; therefore, you also have $\frac1n\leqslant\sqrt{n+1}-\sqrt n$ and so the direct comparison test can be used here.

On the other hand, in the case of the root test, you cannot replace $\lim_{n\to\infty}\sqrt[n]{|a_n|}<1$ by $\lim_{n\to\infty}\sqrt[n]{|a_n|}\leqslant1$; a similar remark applies to the quotient test.

  • $\begingroup$ Ah ok! So showing $<$ implies $\leq$ then. Thank you! $\endgroup$ Commented Feb 12, 2021 at 21:52
  • $\begingroup$ I'm glad I could help. $\endgroup$ Commented Feb 12, 2021 at 22:00

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