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.
 A: 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 $n\in\mathbb{N}.$
A: 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:
$$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}}> 2\sum_{n=1}^{N}\left(\sqrt{n+1}-\sqrt{n}\right) = 2\sqrt{N+1}-2.$$
A: For the induction step, you need to show that
$$2\sqrt{x}+\frac1{\sqrt{x+1}}< 2\sqrt{x+1}$$
$$\frac1{\sqrt{x+1}}< 2(\sqrt{x+1}-\sqrt x)$$
Multiplying both sides by $\sqrt{x+1}+\sqrt{x}$, we have:
$$\iff \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}}< 2 $$
$$\iff \frac{\sqrt x}{\sqrt{x+1}}< 1$$
which is now obvious.
A: $$\sum^n_{k=1}k^{-\frac 1 2}<\int_0^nx^{-\frac 1 2}\ \text{d}x=2n^{\frac 1 2}$$
Since the first value is a lower Riemann sum for the integral in the middle.
