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Why does $\sum_{n=1}^\infty \sqrt{n+1}-\sqrt{n}$ diverge?

Using the ratio test I get the following. First of all since $u_n=\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$

Then $\left| \frac{u_{n+1}}{u_n} \right|=\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}}\lt 1$.

By the ratio test, the series should converge, but it does not. What am I doing wrong?

eager2learn
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    That ratio may be less than $1$, but its limit equals $1$, so the Ratio Test is inconclusive. – 2'5 9'2 Jul 31 '14 at 08:49
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    Because $$s_n =\sum_{k=1}^n (\sqrt{k+1}-\sqrt{k} )=\sqrt{n+1} -1\to\infty$$ –  Jul 31 '14 at 08:50
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    for one: $ \frac{1}{\sqrt{n+1} + \sqrt{n}} > \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{1}{2\sqrt{n}}$ which is known to diverge since it is a p series with power constant less than 1 – Sidharth Ghoshal Jul 31 '14 at 08:50
  • Thanks guys, I forgot that the ratio needs to be less than 1 for all n greater than some $n_0$. – eager2learn Jul 31 '14 at 08:53
  • @frogeyedpeas I think your inequality is wrong, probably meant to have some $+1$'s – DanZimm Jul 31 '14 at 09:14
  • "Thanks guys...": No, you have missed the point. In this case, the ratio is less than $1$ for all $n$ greater than some $n_0$ $-$ for instance, $n_0 = 1$. But the ratio tends to $1$ as $n$ tends to $\infty$, so the ratio test is inconclusive. – TonyK Jul 31 '14 at 09:32
  • @TonyK Well but if the ratio converges to 1, then by definition there exists an $n_0 : \forall n \geq n_0 : \left| \left| \frac{a_n+1}{a_n} \right| -1 \right | \lt \epsilon$. So how can the ratio be less than 1 for all $n\in \mathbb{N}$? Doesn't the limit imply that they are essentially the same given a large enough value for n? – eager2learn Jul 31 '14 at 11:36
  • Your understanding of the ratio test seems to be at fault. What is the ratio test, in your view? – TonyK Jul 31 '14 at 11:39
  • Quoting from my lecture notes: "Let $(a_k){k \in \mathbb{N}}$ be a sequence of real numbers with $a_k \neq 0, \forall k \geq \hat{k} \in \mathbb{N}$. If there exists a $q \in (0,1)$ and a $k_0 \geq \hat{k}: \left|\frac{a{k+1}}{a_k} \right| \leq q, \forall k \geq k_0$ then $\sum a_k$ converges absolutely." – eager2learn Jul 31 '14 at 11:54
  • Yes, exactly. What is your $q$? – TonyK Jul 31 '14 at 12:22
  • Ok I think you can't pick one, since for any $q \in (0,1)$ there would come an $n \in \mathbb{N}$ for which q would be smaller than that value. Is that the reason why you cannot use the ratio test here? – eager2learn Jul 31 '14 at 12:25
  • Yes, exactly!${}$ – TonyK Jul 31 '14 at 14:11

2 Answers2

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The ratio test actually doesn't conclude anything. Note that $$ \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}} = \frac{\sqrt{1 + 1/n} + 1}{\sqrt{1+2/n}+\sqrt{1+1/n}} \to \frac{1+1}{1+1} = 1 $$

Now notice that for large $n$ we have $$ n > 2\sqrt{n+1} > \sqrt{n+1} + \sqrt{n} \implies \frac{1}{n} < \frac{1}{\sqrt{n+1}+\sqrt{n}} $$ So we get that the series diverges.

DanZimm
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First we write the summand as

$$ a_n=(\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}} \sim_{n\to \infty} \frac{1}{2\sqrt{n}}=b_n, $$

then we use the comparison test:

Suppose $\sum_{n} a_n$ and $\sum_n b_n $ are series with positive terms, then

if $\lim_{n\to \infty} \frac{a_n}{b_n}=c>0$, then either both series converge or diverge.

Community
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Mhenni Benghorbal
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