1. ## induction

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.

2. You now need to prove

1/sqrt(n+1) < 2*sqrt(n+1)-2*sqrt(n)

not the one written above

3. ## still dont get it

I guess i dont know why I have to substract 2sqrt(n) from 2sqrt(n+1).
nor do can i continue the induction after that
thanks

4. 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.

5. Use $\displaystyle \frac{1}{\sqrt{n}}=\frac{2}{2\sqrt{n}}<\frac{2}{\s qrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$.

Then when you sum you will have a telescoping series.