
Can you help me with any of these?
 For any natural number n > 1,
prove that
(4^{n}) / (n + 1) < [(2n)!] / [(n!)^{2}].
 For any natural number n > 1, prove that
1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) > sqrt(n).
 For any natural number n and any x > 0, prove that
x^{n} + x^{n  2} + x^{n  4} + ... + x^{n} >= n + 1.
Thanks,
John
10th Grader
Hi John,
I'll do problem 2.
You first ned to verify that the statement is true when n = 2.
1/sqrt(1) + 1/sqrt(2) = 1 + 1/sqrt(2) > 1.7 > sqrt(2)
Now assume that the statement is true for some integer k which is greater than 2. That is
1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) > sqrt(k). *
You are assuming that this statement is true for the specific integer k. What you need to do is to show that
1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) + 1/sqrt(k+1) > sqrt(k+1)
is also true. You can use the asumption * above to see that
1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) + 1/sqrt(k+1)
> sqrt(k) + 1/sqrt(k+1)
= (sqrt(k)sqrt(k+1) + 1)/sqrt(k+1)
= (sqrt(k(k+1)) + 1)/sqrt(k+1)
= (sqrt(k^{2} + k) + 1)/sqrt(k+1)
> (sqrt(k^{2}) + 1)/sqrt(k+1)
= (k+1)/sqrt(k+1)
= sqrt(k+1)
This completes the induction step and hence
1/sqrt(3) + ... + 1/sqrt(n) > sqrt(n)
is true for all n > 1.
Cheers,
Penny
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