does anyone have any ideas on this proof. I have been stuck on it for hours now. Thanks for the help in advance Image of Question is in this link
Prove that for all natural numbers $\geq 1$:
$$\frac {1}{\sqrt1} +\frac {1}{\sqrt2}+\frac {1}{\sqrt3}+.....+\frac {1}{\sqrt n} \leq 2 \sqrt n $$
On
If a non-inductive proof is acceptable, note that for any decreasing function $f(x)$, $$\sum_{x=1}^n f(x) \le \int_0^n f(x) \;dx $$ (If you have trouble seeing this, it might help to make a sketch.)
Taking $f(x) = x^{-1/2}$, we have $$\sum_{x=1}^n x^{-1/2} \le \int_0^n x^{-1/2} \; dx = 2 n ^{1/2}$$
Basis $$\frac{1}{\sqrt{1}}<2\sqrt{1}$$
Induction hypothesis $$\sum_{k=1}^n\frac{1}{\sqrt{k}}<2\sqrt{n}$$
Inductive step To Show: $$\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}}<2\sqrt{n+1}$$
$$LHS=\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}}$$ $$<2\sqrt{n}+\frac{1}{\sqrt{n+1}}$$ $$=2\sqrt{n+1}-(2\sqrt{n+1}-2\sqrt{n}-\frac{1}{\sqrt{n+1}})$$ $$=2\sqrt{n+1}-(\frac{2}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n+1}})$$ $$<2\sqrt{n+1}=RHS$$