Proving that the series 1 + ... + $1 / \sqrt{x}$ < $2 \sqrt{x}$ Proving that the series 1 + ... + $1 / \sqrt{x}$ < $2 \sqrt{x}$
I am doing it by simple induction adding
$1/\sqrt{x+1}$ to both sides, but I can't find a way to manipulate this expression and find that the new series is $< 2 \sqrt{x+1}$.
Can someone show me the correct process?
I failed many times and the furthest I've gotten has been to prove that it's $< 2 \sqrt{x+1} + 1$ which is close, but not enough.

I suppose $x$ is an integer and you series is 
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{n}}$$
For $n=1,$ our inequity is obviously true.
Suppose it is true for $n=k.$ Then 
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{k}}<2\sqrt{k}$$ 
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}<2\sqrt{k}+\dfrac{1}{\sqrt{k+1}}$$  Note that 
$$\sqrt{k+1}-\sqrt{k}=\dfrac{1}{\sqrt{k+1}+\sqrt{k}}>\dfrac{1}{2\sqrt{k+1}}.$$ Hence $$\dfrac{1}{2\sqrt{k+1}}+\sqrt{k}<\sqrt{k+1}$$ and therefore 
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}<2\sqrt{k+1}$$ Therefore by MI, our result is true for all $n\in\mathbb{N}.$
