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I need to determine if the series $$\sum_{n=1}^{\infty} \frac{1}{\sqrt n + \sqrt{n+1}}$$ converges or diverges.

My work so far:

Using the comparison test, $\frac{1}{\sqrt n + \sqrt{n+1}} < \frac{1}{\sqrt n } $ and $\frac {1}{\sqrt n }$ is a divergent series

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  • $\begingroup$ It is a telescopic series, and a clearly divergent one. $\endgroup$ – Jack D'Aurizio Jan 18 '17 at 0:48
  • $\begingroup$ @JackD'Aurizio what is that exactly? $\endgroup$ – bjp409 Jan 18 '17 at 0:49
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    $\begingroup$ If the $n$-th term of a sequence can be written as $a_n = b_{n+1}-b{n}$, then the series $\sum_{n\geq 1}a_n$ is said telescopic and $$\sum_{n=1}^{N}a_n = b_{N+1}-b_{1}.$$ In the present case, $b_n=\sqrt{n}$ is clearly unbounded, hence the series is divergent. And almost as obviously, you cannot use an argument like $\sum_{n\geq 1}a_n \leq \text{(a divergent series)}$ to state something interesting about the LHS. $\endgroup$ – Jack D'Aurizio Jan 18 '17 at 0:52
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There is no point of dominating a given sequence by a divergent sequence to show divergence. Conclusions can be drawn in the following cases:

1) The function dominates (loosely, is greater than) a divergent sequence, in which case it is divergent.

2) The function is dominated by (loosely, is smaller than) a convergent sequence, in which case it is convergent.

This question should be done as follows: Note that $\frac{1}{\sqrt n + \sqrt{n+1}} = \frac{\sqrt {n+1} - \sqrt n}{(n+1)-n} = \sqrt{n+1}-\sqrt{n}$, and that the sum $\sum \sqrt{n+1}-\sqrt n$ is divergent, hence the original sum is divergent.

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  • $\begingroup$ could you explain the note that part, just how the first equation equals the second and then the third $\endgroup$ – bjp409 Jan 18 '17 at 0:44
  • $\begingroup$ We are given something like $\frac{1}{\sqrt a + \sqrt b}$, right? Now, mutiply top and bottom by $\sqrt a - \sqrt b$, and you get $\frac{\sqrt a - \sqrt b}{(\sqrt a + \sqrt b)(\sqrt a - \sqrt b)}$. Expand the denominator, you will get $\frac{\sqrt a - \sqrt b}{a-b}$. In our case, $a = n+1$, $b=n$, and substituting gives you the "note that" part. $\endgroup$ – Teresa Lisbon Jan 18 '17 at 0:51
  • $\begingroup$ thank you very much, one more question. how did you know that the final sum is divergent? just because both go to infinity? $\endgroup$ – bjp409 Jan 18 '17 at 0:52
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    $\begingroup$ Look at it like this: summing the first few terms, we have: $(\sqrt 2 - \sqrt 1) + (\sqrt 3 - \sqrt 2) + (\sqrt 4 - \sqrt 3) + \ldots$. Can you see that terms are getting cancelled out? What remains at the end is $\sqrt n$, which goes to infinity. This is called a telescoping series, you will encounter many of these. $\endgroup$ – Teresa Lisbon Jan 18 '17 at 0:56
  • $\begingroup$ okay that makes perfect sense, thanks very much $\endgroup$ – bjp409 Jan 18 '17 at 0:57
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It makes no sense to bound above by a divergent series. Instead, use this:

$$\frac1{\sqrt n+\sqrt{n+1}}>\frac1{\sqrt{n+1}+\sqrt{n+1}}=\frac1{2\sqrt{n+1}}$$

which is then appropriate to show divergence since it is a lower bound.

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