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