Too many Fractions and too much Infinity!

Ah. I'm writing a solution after so long. Ok, so here we go,

You will need to know the identity $1+2+3+...+n=\frac{n(n+1)}{2}$ and the formula for calculating the sum of an infinite geometric series . I'll assume that you are familiar with limits and summations in general.

Let's do the problem part by part. First, we will evaluate the limit.

$\displaystyle \lim_{x\to (+\infty)} (\frac{1+2+3+\ldots +x}{x^2}) \\ = \displaystyle \lim_{x\to (+\infty)} (\frac{x(x+1)}{2x^2}) \\ = \displaystyle \frac{1}{2}\centerdot \lim_{x\to (+\infty)} (\frac{x+1}{x}) \\ = \displaystyle \frac{1}{2}\centerdot \lim_{x\to (+\infty)} (1+\frac{1}{x})$

Now, let us consider $z=\frac{1}{x}$ . So, $x\to (+\infty) \implies \frac{1}{x} \to 0 \implies z\to 0$ .

So, we can rewrite the limit as ---->

$\displaystyle \frac{1}{2}\centerdot \lim_{z\to 0}(1+z) = \frac{1}{2} (1+0) = \boxed{\frac{1}{2}}$

Now, putting the value of this limit in the given problem we can express the given problem as ---->

$\displaystyle \frac{3}{5}\centerdot \sum_{n=4}^\infty \bigg( \frac{1}{2}\centerdot \frac{1}{2^n} \bigg) = \frac{a}{b}$

$\implies \displaystyle \frac{3}{5}\centerdot \sum_{n=4}^\infty \bigg( \frac{1}{2^{n+1}} \bigg)=\frac{a}{b}$

$\implies \displaystyle \frac{3}{5} (\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}+ \ldots)=\frac{a}{b}$

$\displaystyle \implies \frac{3}{5}\bigg( \dfrac{\frac{1}{2^5}}{1-\frac{1}{2}} \bigg)=\frac{a}{b}$

$\implies \frac{3}{5}\centerdot \frac{1}{2^4}=\frac{a}{b}$

$\implies \frac{a}{b}=\frac{3}{80}$

We can clearly see that the fraction is in its simplest form and $\gcd(3,80)=1$ , so we have $a=3\text{ and } b=80$ .

Now, the required value $=ab-1=3\times 80 -1 =240-1=\boxed{239}$