SOLUTION: Prove that {{{6/(n+1) <= 6/(2n+1) +sqrt(sum(1/k^2, k=1,n))}}} for {{{n >= 1}}}.

Algebra ->  Sequences-and-series -> SOLUTION: Prove that {{{6/(n+1) <= 6/(2n+1) +sqrt(sum(1/k^2, k=1,n))}}} for {{{n >= 1}}}.      Log On


   

Question 1062031: Prove that
6%2F%28n%2B1%29+%3C=+6%2F%282n%2B1%29+%2Bsqrt%28sum%281%2Fk%5E2%2C+k=1%2Cn%29%29 for n+%3E=+1.
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Prove that

6%2F%28n%2B1%29+%3C=+6%2F%282n%2B1%29+%2Bsqrt%28sum%281%2Fk%5E2%2C+k=1%2Cn%29%29 for n+%3E=+1

That will be true if and only if

6%2F%28n%2B1%29+-+6%2F%282n%2B1%29++%3C=+sqrt%28sum%281%2Fk%5E2%2C+k=1%2Cn%29%29 for n+%3E=+1

Simplifying the expression on the left,

%286n%29%2F%28%28n%2B1%29%282n%2B1%29%29++%3C=+sqrt%28sum%281%2Fk%5E2%2C+k=1%2Cn%29%29 for n+%3E=+1

The proof is immediate because the left side decreases as
n gets larger and the right side increases as n gets larger,
so the inequality will always hold for n+%3E=+1.

Edwin