# The inequality in applying convergence/divergence tests

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.

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.
• Ah ok! So showing $<$ implies $\leq$ then. Thank you! Feb 12 '21 at 21:52