You now need to prove
1/sqrt(n+1) < 2*sqrt(n+1)-2*sqrt(n)
not the one written above
If you still cannot solve this problem, please ask again.
Show that for all positive integers n , we have.
1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) < 2*sqrt(n)
I have shown the base case:
Now i need to assume the inductive case to show
...... 1/sqrt(n+1) < 2*sqrt(n+1)
and there is when i get stuck.
The induction argument should go like this:
Assuming
1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) < 2*sqrt(n) (1) for some n
We now prove
1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) + 1/sqrt(n+1) < 2*sqrt(n+1) (2)
Now if we can prove
1/sqrt(n+1) < 2*sqrt(n+1)-2*sqrt(n) (3)
When you add up (1) and (3) together, you will get (2).
To prove (3), you can multiply both sides by sqrt(n+1), then complete the square.