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 :)
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,
A very short (though non-inductive) proof:
$$ \sum_{k=1}^n \frac{1}{\sqrt{k}} \ge \sum_{k=1}^n \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sum_{k=1}^n \frac{\sqrt{k} - \sqrt{k-1}}{(\sqrt{k} + \sqrt{k-1})(\sqrt{k} - \sqrt{k-1})} = \sum_{k=1}^n (\sqrt{k} - \sqrt{k-1}) = \sqrt{n} $$
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.
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.
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.
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 $$ \sum_{i=1}^n \frac 1{\sqrt k} \ge \sum_{i=1}^n \frac 1{\sqrt n} = \frac n{\sqrt n} = \sqrt n. $$
