# Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction

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 :)

• You are going to have lots of problems proving that equality, for it is false. (Consider, for example, the case where $n=2$) – Mariano Suárez-Álvarez Aug 8 '11 at 16:23
• 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? – Patrick Da Silva Aug 8 '11 at 16:24
• 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. – Arturo Magidin Aug 8 '11 at 16:24
• 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. – user9413 Aug 8 '11 at 16:26
• Or maybe he just wants to prove $\ge$ – GEdgar Aug 8 '11 at 16:26

## 6 Answers

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,

• 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. – Mariano Suárez-Álvarez Aug 8 '11 at 16:31
• (After your edit, the same problem subsists: now «this is equivaent to $X\iff Y$» :) ) – Mariano Suárez-Álvarez Aug 8 '11 at 16:33
• I was editing a lot : are there any complaints now? =) – Patrick Da Silva Aug 8 '11 at 16:37
• I did not understand the last step could you please elaborate? – Jason Aug 8 '11 at 16:39
• 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? – Patrick Da Silva Aug 8 '11 at 16:43

A very short (though non-inductive) proof:

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

• Nice! Of course there is some hidden induction. [I'd put the $\sqrt k-\sqrt{k-1}$ in parenthesis in the last sum.] – Pierre-Yves Gaillard Aug 8 '11 at 18:42

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.

• Couldn't you just write "greater than $n/\sqrt{n} = \sqrt n$" instead of keeping the $1/\sqrt{1}$ aside? – Patrick Da Silva Aug 8 '11 at 16:33
• By the way, you can write \ge for $\ge$. – Patrick Da Silva Aug 8 '11 at 16:36
• Thanks for $\ge$. Don't get your question right. You mean $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}\ge \frac{n}{\sqrt{n}}=\sqrt{n}$$? – ulead86 Aug 8 '11 at 16:42
• Exactly. Easier, isn't it? – Patrick Da Silva Aug 8 '11 at 17:05
• @DDaniel: The induction is not explicit, though in fact there is an implicit induction hidden in the $\dots$. Upvote for a probably simplest solution! – André Nicolas Aug 8 '11 at 17:05

You know that $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$, and your goal is to show that $\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}\ge\sqrt{n+1}}$. 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.

• Great thanks, this is the solution I got to Though my algebra proof was different – Jason Aug 8 '11 at 19:56

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.

• Can you give a characterization of the $n$'s for which is it stronger? =O Kidding. +1 – Patrick Da Silva Aug 8 '11 at 23:24
• @Patrick Da Silva: Any integer bigger than $\frac{16}{9}$ – robjohn Aug 9 '11 at 0:02
• Lollll you guys are so serious. XD Anyway that's cool. – Patrick Da Silva Aug 9 '11 at 1:12
• @Patrick Da Silva: I was joking, too, but maybe my sense of humor is off . o O (integer bigger than $\frac{16}{9}$...) :-p – robjohn Aug 9 '11 at 2:23
• 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. – Patrick Da Silva Aug 9 '11 at 2:24

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 $$\sum_{i=1}^n \frac 1{\sqrt k} \ge \sum_{i=1}^n \frac 1{\sqrt n} = \frac n{\sqrt n} = \sqrt n.$$

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