Prove a Sum is Larger than a Root

  • Thread starter Thread starter hammonjj
  • Start date Start date
  • Tags Tags
    Root Sum
hammonjj
Messages
32
Reaction score
0

Homework Statement


Let nεZ>0. Prove that:

Ʃ1/√(i) ≥ √(n)

The Attempt at a Solution


I'm not sure where to begin, this feels like it should be an induction problem, but I'm not sure how to show that √(n) + 1/√(i+1) ≥ √(n+1). There doesn't seem to be any obvious algebra that would simplify this into what I need it to be.
 
Physics news on Phys.org
Statement correct for n=1.
For ##n\ge2,##
\sum_{i=1}^n\frac1{\sqrt i}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx<br /> =2\sqrt{n+1}-2\ge\sqrt n.
 
No, you have to show that:

√(n) + 1/√(n+1) ≥ √(n+1).

In your last step of the induction proces.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top