# Square root series convergence or divergence?

I'm trying to show whether the series $$\sum_{n=1}^{\infty} \sqrt{n+1} - \sqrt{n}$$ converges or diverges using the direct comparison test.

Attempt:

$$\sqrt{n+1} - \sqrt{n} = \sqrt{n}\Big(\sqrt{1+ \frac{1}{n}} - 1\Big)$$.

Then for positive $$t$$ we have $$1 < \sqrt{1+t} < 1 + \frac{t}{2}$$, because we have $$(1 + \frac{t}{2})^{2} > 1 + t$$.

Therefore we have $$0 < \sqrt{n}\Big(\sqrt{1 + \frac{1}{n}} - 1\Big) < \frac{1}{2 \sqrt{n}}$$.

Now I know $$\frac{1}{2 \sqrt{n}}$$ diverges. But I'm not sure I can claim the original series diverges since for the Direct comparison test to apply I have to show the original series is greater than a divergent series?

Thanks.

• Isn't $$\sum_{k=1}^n \sqrt {k+1} - \sqrt k = \sqrt {n+1} - 1$$ ?? Nov 16, 2019 at 10:56

## 3 Answers

Directly with a known test:

$$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\ge\frac1{\sqrt{2n}+\sqrt n}=\frac1{\sqrt2+1}\cdot\frac1{\sqrt n}$$

and the last one is a sequence whose infinite series diverges as a scalar multiple of the divergent series $$\;\sum\limits_{n=1}^\infty\cfrac1{\sqrt n}\;$$ ...and now use the comparison test for positive series.

Another way: observe that

$$S_k:=\sum_{n=1}^k(\sqrt{n+1}-\sqrt n)=\left(\sqrt2-\sqrt1\right)+\left(\sqrt3-\sqrt2\right)+\ldots+\left(\sqrt{k+1}-\sqrt k\right)=$$

$$=\sqrt{k+1}-1\xrightarrow[k\to\infty]{}\infty$$

and this means the sequence of partial sums of the series diverges to $$\;\infty\;$$ and thus so does the infinite series.

We can use that

$$\sqrt{n}\left(\sqrt{1+ \frac{1}{n}} - 1\right) \sim \sqrt{n} \cdot\frac1{2n}=\frac1{2\sqrt n}$$

and then conclude by limit comparison test.

As an alternative we can use that

$$\sqrt{1+ \frac{1}{n}} \ge 1+ \frac{1}{2n}-\frac{1}{8n^2}$$

and conclude by direct comparison test.

• I haven't covered the limit comparison test yet, so was wondering If i can conclude someway using direct comparison test or p-test. I think I'm nearly there except for the above issue. Nov 16, 2019 at 10:58
• @AnalysisLearner Take a look to it because it allows to solve many problem in a simpler way.
– user
Nov 16, 2019 at 10:59
• @AnalysisLearner I add an alternative by direct comparison test.
– user
Nov 16, 2019 at 11:00
• Thanks I'll take a look. I keep seeing everyone posting this $\sim$ sign. What does it means? Nov 16, 2019 at 11:08
• @AnalysisLearner We have $a_n=\sqrt{n + 1} - \sqrt{n}$ and $b_n=\frac{1}{2 \sqrt{n}}$ then $$\frac{a_n}{b_n}=\frac{\sqrt{n + 1} - \sqrt{n}}{\frac{1}{2 \sqrt{n}}} \to \frac12$$
– user
Nov 16, 2019 at 12:18

Actually, we have $$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\sim_{n\to\infty}\frac1{2\sqrt n}$$ and the latter diverges, so the former does too.

Added:

I'll show why it two series $$\sum_n a_n$$ and $$\sum_n b_n$$ have asymptotically equivalent (ultimately) positive terms, and if one of them diverges, the other diverges too.

Indeed,suppose $$\sum_n b_n$$ diverges. By definition of equivalence, $$\lim_{n\to\infty}\dfrac{a_n}{b_n}=1$$, so for any $$\varepsilon>0$$, there exists an integer $$N_0$$ such that \begin{align} &&1-\varepsilon<\frac{a_n}{b_n}&<1+\varepsilon&&\text{ for all }\quad n\ge N_0\qquad \\ &\text{whence, as }\; b_n>0, &a_n&>(1-\varepsilon)b_n&&\text{ for all }\quad n\ge N_0, \end{align} which by the comparison test, implies that $$\sum_n a_n$$ diverges, just like the series $$(1-\varepsilon)\sum_n b_n$$.

• why can we claim that the former diverges just because the latter does? What test is this using? As you see, I get to the same value of $\frac{1}{2\sqrt{n}}$ but I wasn't sure if/why I could claim divergence of the former... Nov 16, 2019 at 11:03
• @AnalysisLearner: I've added a proof of divergence. Is that clearer? Nov 16, 2019 at 19:09