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!
Best Answer
You have proved it; you just need to write the 6 inequalities in the reverse order, and connect them by implication $\implies$ or keep the ordering and show that each is equivalent $\iff$ with the next.