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Math Induction..?
I have no idea how to do this problem. The less than symbol being before the sequence is throwing me for a loop.
sqrt(n) < 1/sqrt(1) + 1/sqrt(2)+ ... + 1/sqrt(n), for all integers n >= 2.
Thanks for any help.
3 Answers
- 1 decade agoFavourite answer
1. Is it true for n = 2 (the initial value of n)?
1/sqrt(1) + 1/sqrt(2)
= 1.707106...
> sqrt(2)
True
2. Assume true for some k >= 2.
1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k) > sqrt(k)
3. Prove true for (k+1)
1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k+1) >? sqrt(k+1)
1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k) + 1/sqrt(k+1) >? sqrt(k+1) [expansion]
sqrt(k) + 1/sqrt(k+1) >? sqrt(k+1) [since true for some k>=2]
sqrt(k)sqrt(k+1) + 1 >? k+1 [multiply both sides by sqrt(k+1)]
sqrt(k)sqrt(k+1) >? k [subtract 1 from both sides]
k(k+1) >? k^2 [square both sides]
k^2 + k >? k^2...True since k>=2
- 1 decade ago
1) if n=2, then sqrt(n)=sqrt(2)=sqrt(2)/2+sqrt(2)/2 and 1/sqrt(1)+1/sqrt(2)=1+sqrt(2)/2. so the proposition is correct.
2) suppose sqrt(n)<1/sqrt(1) + 1/sqrt(2)+ ... + 1/sqrt(k).
when n=k+1, 1/sqrt(1) + 1/sqrt(2)+ ... + 1/sqrt(k+1)>sqrt(k)+1/sqrt(k+1)=sqrt(k+1)((sqrt(k)sqrt(k+1)+1)/sqrt(k+1))>sqrt(k+1)((k+1)/(k+1))=sqrt(k+1)
so the proposition is correct for n=k+1.
so the proposition is correct for all integers n>=2
