Summation of inverse of square-roots of odd numbers

Question

I need to find $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} +...+\frac{1}{\sqrt{99}}$$. I have learnt how to sum up a consecutive order like this: $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +...+\frac{1}{\sqrt{1000000}}$$ but I am unfamiliar with the one comprising of odd numbers. Any help please? For this, I would prefer if integration is not used.

Answer 1

If sums of consecutive terms of $1/\sqrt{n}$ are not a problem (?) then note that $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} +\dots+\frac{1}{\sqrt{99}} =\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+\frac{1}{\sqrt{100}} -\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+\frac{1}{\sqrt{50}}\right).$$

Answer 2

Hint: You should take an interest in the sum of EVEN numbers, and notice that the sum of all numbers can be split in even and odd numbers. By soustraction you will have the sum of odd numbers.