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.

Solution 1:

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 $n\in\mathbb{N}.$

Solution 2:

We can notice that: $$ \sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}\in\left(\frac{1}{2\sqrt{n+1}},\frac{1}{2\sqrt{n}}\right), $$ hence: $$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}}< 2\sum_{n=1}^{N}\left(\sqrt{n}-\sqrt{n-1}\right) = 2\sqrt{N}$$ as wanted. On the other hand, we also have: $$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}}> 2\sum_{n=1}^{N}\left(\sqrt{n+1}-\sqrt{n}\right) = 2\sqrt{N+1}-2.$$