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With the advise submitted by Thomas Andrews the written form isn't any thing except a Riemann sum for calculating the integral $$\int_{0}^1 \left\{\frac{1}{x}\right\}\,dx.$$ Thus we should calulate this integral. we have: $$\int_{0}^1 \left\{\frac{1}{x}\right\}\,dx = \sum _{n=1}^\infty \int_{\frac{1}{n+1} }^{\frac{1}{n} } \left\{\frac{1}{x}\right\}\,dx ... 0 I'm assuming you're only working with real numbers. In that case, to be taking square roots you have to have x > 0 and y \geq 0. To solve this problem, let's fix x, and then find all y such that the given thing converges. So given x, let z = \sqrt{x}, so$$e^n \frac{(\sqrt{y} - \sqrt{x})^{2n}}{x^n} = \sqrt{e}^{2n} \frac{(\sqrt{y} - ...
You're on the right track. As mentioned in the comments you don't need a "$+C$" on both sides of the equation. Each integral contributes an additive constant, the two of which we can combine into a single constant. Now you just need to solve for $y$. Start by isolating the only term with a $y$ in it. $$\frac{-1}{2} e^{-2y} = \frac{1}{3}e^{3x}+C$$  ...