Saw this challenge problem:
$$\frac{\sum_{n=0}^{n=\infty} \frac{1}{\sqrt{3n+1}} - \frac{1}{\sqrt{3n+2}}}{\sum_{n=0}^{n=\infty} \frac{1}{\sqrt{6n+1}} - \frac{1}{\sqrt{6n+5}}} = 2 - \sqrt2$$
How to tackle this? Tried getting partial sums to telescope and rationalizing numerator/denominator but didn't get anywhere. I noticed that the denominator terms are contained within the numerator terms.
Let the numerator and denominator be $N$ and $D$, respectively. Writing the even and odd terms together in $N$, $$N = \sum_{n=0}^{\infty} \left ( \frac{1}{\sqrt{6n+1}} -\frac{1}{\sqrt{6n+2}} +\frac{1}{\sqrt{6n+4}} -\frac{1}{\sqrt{6n+5}} \right) \\ = D -\sum_{n=0}^{\infty} \frac{1}{\sqrt{6n+2}} -\frac{1}{\sqrt{6n+4}} \\ =D-\frac{1}{\sqrt 2} N \\ \implies \frac ND = \frac{\sqrt 2}{\sqrt 2+1} =2-\sqrt 2 $$