Prove a Sum is Larger than a Root

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SUMMARY

The discussion centers on proving the inequality Ʃ1/√(i) ≥ √(n) for n in the set of positive integers. Participants suggest using mathematical induction to demonstrate that √(n) + 1/√(n+1) ≥ √(n+1) holds true. The proof involves comparing the sum of the series to an integral, specifically ∫1^(n+1) 1/√(x) dx, which simplifies to 2√(n+1) - 2. The conclusion confirms the validity of the statement for n ≥ 2, with a base case established for n = 1.

PREREQUISITES
  • Understanding of mathematical induction
  • Familiarity with inequalities and their proofs
  • Basic knowledge of integrals, specifically ∫1^(n+1) 1/√(x) dx
  • Proficiency in manipulating square roots and summations
NEXT STEPS
  • Study mathematical induction techniques in depth
  • Explore the properties of integrals and their applications in inequalities
  • Learn about series convergence and divergence
  • Investigate advanced inequality proofs, such as Cauchy-Schwarz and Jensen's inequality
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Students in mathematics, particularly those studying calculus and analysis, educators teaching proof techniques, and anyone interested in advanced mathematical inequalities.

hammonjj
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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.
 
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Statement correct for n=1.
For ##n\ge2,##
[tex]\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.[/tex]
 
No, you have to show that:

√(n) + 1/√(n+1) ≥ √(n+1).

In your last step of the induction proces.
 

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