<s>[INST] Induction: show that $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$ The question:
Induction: show that: 
$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
for $n \geq 1$
My attempt at a solution:
First we test the base case: $n=1$ this gives us:
$$1 < 2$$
Which works.
Then we do the inductive assumption that it holds true for $k=n$, this gives us:
$$1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
If we can prove that the following is true, we have solved the problem: 
$$1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n+1}} < 2\sqrt{n+1}$$
we then have to show that $$2\sqrt{n} + \frac{1}{\sqrt{n+1}} \leq 2\sqrt{n+1}$$
Which can be written as: $$2\sqrt{n}\sqrt{n+1} + 1 \leq 2(n+1)$$
$$1 \leq 2(n+1) - 2\sqrt{n}\sqrt{n+1}$$
$$2\sqrt{n}\sqrt{n+1} \leq 2(n+1) - 1$$
$$2\sqrt{n}\sqrt{n+1} \leq 2n+1$$
$$(2\sqrt{n}\sqrt{n+1})^2 \leq (2n+1)^2$$
$$4n^2 + 4n \leq 4n^2 + 4n + 1$$
Can anyone please help me to continue from here or does this mean that I have proved the inductive assumption?
Thank you!
 [/INST] You have proved it but if you dislike switching to (essentially) showing a new inequality is true you could approach this slightly differently. 
$\begin{align}
1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n+1}} 
&< 2\sqrt{n} + \frac{1}{\sqrt{n+1}} \mathrm{(by\ inductive\ hypothesis)} \\
&= \frac{2\sqrt{n}(n+1)}{n+1} + \frac{\sqrt{n+1}}{n+1} \\
&= \frac{(2\sqrt{n}\sqrt{n+1}+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2\sqrt{n^2+n}+1)\sqrt{n+1}}{n+1} \\
&< \frac{(2\sqrt{n^2+n+\frac{1}{4}}+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2\sqrt{(n+\frac{1}{2})^2}+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2(n+\frac{1}{2})+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2n+1+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2n+2)\sqrt{n+1}}{n+1} \\
&= \frac{2(n+1)\sqrt{n+1}}{n+1} \\
&= 2\sqrt{n+1} \\
\end{align}$
</s>