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I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.

$$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$

Perhaps someone can help me out I don't understand how to move forward from here: $$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$ proof and explanation would greatly be appreciated :) Thanks :)

EDIT sorry meant GE not = fixed :)

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    $\begingroup$ You are going to have lots of problems proving that equality, for it is false. (Consider, for example, the case where $n=2$) $\endgroup$ – Mariano Suárez-Álvarez Aug 8 '11 at 16:23
  • $\begingroup$ There is a typo somewhere : your first mathematical line is clearly wrong since $1 + 1/\sqrt{2} > 2$. Did you wish to prove the inequality, as your second line suggests? $\endgroup$ – Patrick Da Silva Aug 8 '11 at 16:24
  • $\begingroup$ The equality is false. If it were true, then $1 + \frac{1}{\sqrt{2}}$ would equal $\sqrt{2}$. But $1+\frac{1}{\sqrt{2}} = \frac{2+\sqrt{2}}{2}$. If this were equal to $\sqrt{2}$, then you would have $2+\sqrt{2}=2\sqrt{2}$, or $2=\sqrt{2}$, which is patently false. $\endgroup$ – Arturo Magidin Aug 8 '11 at 16:24
  • $\begingroup$ You can use the $\mathsf{Euler's \ summation \ formula}$. But as observed by in the previous comments, even I don't think that this is true. $\endgroup$ – user9413 Aug 8 '11 at 16:26
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    $\begingroup$ Or maybe he just wants to prove $\ge$ $\endgroup$ – GEdgar Aug 8 '11 at 16:26
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If you wanted to prove that $$ \sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n, $$ that I can do. It is clear for $n=1$ (since we have equality then), so that it suffices to verify that $$ \sum_{k=1}^{n+1} \frac 1{\sqrt k} \ge \sqrt{n+1} $$ but this is equivalent to $$ \sum_{k=1}^{n} \frac 1{\sqrt k} + \frac 1{\sqrt{n+1}} \ge \sqrt{n+1} \ $$ and again equivalent to $$ \sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} + 1 \ge n+1 $$ so we only need to prove the last statement now, using induction hypothesis. Since $$ \sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n, $$ we have $$ \sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} \ge \sqrt{n+1}\sqrt{n} \ge \sqrt{n} \sqrt{n} = n. $$ Adding the $1$'s on both sides we get what we wanted.

Hope that helps,

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    $\begingroup$ Stylistic complaint: you end with «but this is $X\iff Y$», where you meant «but this is (equivalent to) $X$ which holds since $Y$» or something like that. As you wrote it, it appears that the inequality in your second equation is the same thing as a logical equivalence between two inecualities: this does not make much sense. $\endgroup$ – Mariano Suárez-Álvarez Aug 8 '11 at 16:31
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    $\begingroup$ (After your edit, the same problem subsists: now «this is equivaent to $X\iff Y$» :) ) $\endgroup$ – Mariano Suárez-Álvarez Aug 8 '11 at 16:33
  • $\begingroup$ I was editing a lot : are there any complaints now? =) $\endgroup$ – Patrick Da Silva Aug 8 '11 at 16:37
  • $\begingroup$ I did not understand the last step could you please elaborate? $\endgroup$ – Jason Aug 8 '11 at 16:39
  • $\begingroup$ I multiplied the induction hypothesis by $\sqrt{n+1}$ to get the first inequality, and for the second inequality, since $n+1 > n$, $\sqrt{n+1} > \sqrt n$, so if you multiply those two by $\sqrt{n}$, you get $\sqrt{n+1}\sqrt{n} > \sqrt{n} \sqrt{n}$. That is $\sqrt{n}^2 = n$. Is that okay? $\endgroup$ – Patrick Da Silva Aug 8 '11 at 16:43
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A very short (though non-inductive) proof:

$$ \sum_{k=1}^n \frac{1}{\sqrt{k}} \ge \sum_{k=1}^n \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sum_{k=1}^n \frac{\sqrt{k} - \sqrt{k-1}}{(\sqrt{k} + \sqrt{k-1})(\sqrt{k} - \sqrt{k-1})} = \sum_{k=1}^n (\sqrt{k} - \sqrt{k-1}) = \sqrt{n} $$

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  • $\begingroup$ Nice! Of course there is some hidden induction. [I'd put the $\sqrt k-\sqrt{k-1}$ in parenthesis in the last sum.] $\endgroup$ – Pierre-Yves Gaillard Aug 8 '11 at 18:42
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I won't use induction:

On the left side you have a sum with $n$ terms, the smallest one is $\frac{1}{\sqrt{n}}$. So you get the inequality:

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}\ge \frac{1}{\sqrt{1}}+(n-1)\frac{1}{\sqrt{n}}=\left(1-\frac{1}{\sqrt{n}}\right)+\sqrt{n}$$

And now you can see easily that the right hand side is larger than $\sqrt{n}$, for all $n>1$.

I hope this helps.

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  • $\begingroup$ Couldn't you just write "greater than $n/\sqrt{n} = \sqrt n$" instead of keeping the $1/\sqrt{1}$ aside? $\endgroup$ – Patrick Da Silva Aug 8 '11 at 16:33
  • $\begingroup$ By the way, you can write \ge for $\ge$. $\endgroup$ – Patrick Da Silva Aug 8 '11 at 16:36
  • $\begingroup$ Thanks for $\ge$. Don't get your question right. You mean $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}\ge \frac{n}{\sqrt{n}}=\sqrt{n}$$? $\endgroup$ – ulead86 Aug 8 '11 at 16:42
  • $\begingroup$ Exactly. Easier, isn't it? $\endgroup$ – Patrick Da Silva Aug 8 '11 at 17:05
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    $\begingroup$ @DDaniel: The induction is not explicit, though in fact there is an implicit induction hidden in the $\dots$. Upvote for a probably simplest solution! $\endgroup$ – André Nicolas Aug 8 '11 at 17:05
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You know that ${\displaystyle \sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}}$, and your goal is to show that ${\displaystyle \sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}\ge\sqrt{n+1}}}$. Observe that $$ \sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}} = \sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}} + {1 \over \sqrt{n+1}}$$ $$\geq \sqrt{n} + {1 \over \sqrt{n+1}}$$ You use the induction hypothesis in the above line. So what you need to show is $$\sqrt{n} + {1 \over \sqrt{n+1}} \geq \sqrt{n+1}$$ At this point you can basically try to fool around with the algebra to get it to work out. One example of this would be to multiply both sides by $\sqrt{n+1}$, getting $$\sqrt{n(n+1)} + 1 \geq n + 1$$ Or equivalently, $$\sqrt{n(n+1)} \geq n$$ Squaring both sides gives $$n^2 + n \geq n^2$$ This last equation is obviously true. To make the argument rigorous, you just observe that these steps are reversible; going in the opposite direction from above takes you from $n^2 + n \geq n^2$ to $\sqrt{n} + {1 \over \sqrt{n+1}} \geq \sqrt{n+1}$.

Some people might not like doing this sort of reversal-of-steps argument, but it does have an advantage that you don't really have to see anything clever to do it; ususally playing around with the algebra enough will eventually lead to something obvious.

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  • $\begingroup$ Great thanks, this is the solution I got to Though my algebra proof was different $\endgroup$ – Jason Aug 8 '11 at 19:56
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Using the same identity that Sasha did, $$ \begin{align} \sqrt{k+1}-\sqrt{k}&=\frac{1}{\sqrt{k+1}+\sqrt{k}}\\ &\le\frac{1}{2\sqrt{k}} \end{align} $$ We can sum and multiply by $2$ to get $$ 2(\sqrt{n+1}-1)\le\sum_{k=1}^n\frac{1}{\sqrt{k}} $$ Which for most $n$ is stronger.

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  • $\begingroup$ Can you give a characterization of the $n$'s for which is it stronger? =O Kidding. +1 $\endgroup$ – Patrick Da Silva Aug 8 '11 at 23:24
  • $\begingroup$ @Patrick Da Silva: Any integer bigger than $\frac{16}{9}$ $\endgroup$ – robjohn Aug 9 '11 at 0:02
  • $\begingroup$ Lollll you guys are so serious. XD Anyway that's cool. $\endgroup$ – Patrick Da Silva Aug 9 '11 at 1:12
  • $\begingroup$ @Patrick Da Silva: I was joking, too, but maybe my sense of humor is off . o O (integer bigger than $\frac{16}{9}$...) :-p $\endgroup$ – robjohn Aug 9 '11 at 2:23
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    $\begingroup$ I know it was a joke, that's why I added the XD in the end. If I had believed you were truly serious I would've been pissed. $\endgroup$ – Patrick Da Silva Aug 9 '11 at 2:24
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For those who strive for non-induction proofs...

Since $\frac 1{\sqrt k} \ge \frac 1{\sqrt n}$ for $1 \le k \le n$, we actually have $$ \sum_{i=1}^n \frac 1{\sqrt k} \ge \sum_{i=1}^n \frac 1{\sqrt n} = \frac n{\sqrt n} = \sqrt n. $$

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    $\begingroup$ This shows how understated the original estimate was. It's like saying that $\sum_{k=1}^n k\le n^2$. $\endgroup$ – robjohn Aug 8 '11 at 23:41
  • $\begingroup$ I know... and it also shows why I'm so impressed that people made a huuuge deal about finding 430423 proofs for this. $\endgroup$ – Patrick Da Silva Aug 8 '11 at 23:42
  • $\begingroup$ @robjohn Talking of understated, I'm usually happy with $\sum \limits_{k=1}^{n} k = O(n^2)$. :-) $\endgroup$ – Srivatsan Jan 4 '12 at 21:58

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