Show that for all $n \in \mathbb{N}$
$$ 2(\sqrt{n+1} -1) \lt \sum_{k=1}^{n}{k^{-1/2}} \lt 2\sqrt{n}$$
I know that I am suppose to use induction and break it up into $$ 2(\sqrt{n+1} -1) \lt \sum_{k=1}^{n}{k^{-1/2}}$$ and $$ \sum_{k=1}^{n}{k^{-1/2}} \lt 2\sqrt{n}$$
But I am having trouble doing this. Can someone provide the full proof?
On
To show the right hand inequality:
Let $S_n$ denote the sum and assume that we have shown $S_n<2\sqrt n$. Then we wish to prove $$S_{n+1}=S_n+\frac 1{\sqrt {n+1}}<2\sqrt {n+1}$$ Inductively we see that it would suffice to prove $$2\sqrt n+\frac 1{\sqrt {n+1}}=\frac {2\sqrt {n(n+1)}+1}{\sqrt {n+1}}<2\sqrt {n+1}$$ Clear the denominator to see that this is equivalent to $$2\sqrt {n(n+1)}+1<2(n+1)\quad \text {or} \quad 2\sqrt {n(n+1)}<2n+1$$ Square both sides to see that this is in turn is equivalent to $$4n^2+4n<4n^2+4n+1$$ Which, finally, is self-evident.
This is probably not the smartest way to turn it, but here is try.
showing $$ 2(\sqrt{n+1}-1)< \sum_{k=0}^{n}{k^{-1/2}}$$ Base step: Showing it hold for n=1, we have $ 2 \sqrt{2}-2 < 1 + \frac{1}{\sqrt{2}} $ which is obviously true.
Induction step, showing that assuming that it is true for n, then it's true for n+1. $$\sum_{k=0}^{n+1}{k^{-1/2}} =(n+1)^{-1/2} + \sum_{k=0}^{n}{k^{-1/2}} $$ now using the assumption. $$2(\sqrt{n+2}-1)* \leq(n+1)^{-1/2} + 2(\sqrt{n+1}-1) < (n+1)^{-1/2} + \sum_{k=0}^{n}{k^{-1/2}}$$
Now showing $$ \sum_{k=0}^{n}{k^{-1/2}} < 2 \sqrt{n} $$
First base step: $$ 1<2$$.
Induction step: $$\sum_{k=0}^{n}{k^{-1/2}} + (n+1)^{-1/2}< 2 \sqrt{n}+(n+1)^{-1/2}<*2\sqrt{n+1} $$
I had a hard time showing*, but one way to do it, would be looking at the continually expansion (for $ n \in \mathbb{R}^+$, and showing that it there is no n, that solves it for equals. Then since it's continually you can just look at one point, but I think there is an easier way