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.


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

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?


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

3 Answers 3


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

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


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.

  • $\begingroup$ 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. $\endgroup$ Nov 16, 2019 at 10:58
  • $\begingroup$ @AnalysisLearner Take a look to it because it allows to solve many problem in a simpler way. $\endgroup$
    – user
    Nov 16, 2019 at 10:59
  • $\begingroup$ @AnalysisLearner I add an alternative by direct comparison test. $\endgroup$
    – user
    Nov 16, 2019 at 11:00
  • $\begingroup$ Thanks I'll take a look. I keep seeing everyone posting this $ \sim$ sign. What does it means? $\endgroup$ Nov 16, 2019 at 11:08
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    $\begingroup$ @AnalysisLearner We have $a_n=\sqrt{n + 1} - \sqrt{n} $ and $b_n=\frac{1}{2 \sqrt{n}}$ then $$\frac{a_n}{b_n}=\frac{\sqrt{n + 1} - \sqrt{n}}{\frac{1}{2 \sqrt{n}}} \to \frac12$$ $\endgroup$
    – 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.


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 $(1-\varepsilon)\sum_n b_n$.

  • $\begingroup$ 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... $\endgroup$ Nov 16, 2019 at 11:03
  • $\begingroup$ @AnalysisLearner: I've added a proof of divergence. Is that clearer? $\endgroup$
    – Bernard
    Nov 16, 2019 at 19:09

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