Can you help me with any of these?

  1. For any natural number n > 1, prove that

    (4n) / (n + 1) < [(2n)!] / [(n!)2].

  2. For any natural number n > 1, prove that

    1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) > sqrt(n).

  3. For any natural number n and any x > 0, prove that

    xn + xn - 2 + xn - 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(k2 + k) + 1)/sqrt(k+1)

> (sqrt(k2) + 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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