we know that $$\sum_{i=1}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$or $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ and we can prove this by telescoping series $$ (i+1)^2-i^2=2i+1$$ $$(i+1)^3-i^3=3i^2+3i+1$$
But while solving some problems, I suddenly thought about $\sum_{i=1}^n \sqrt i$ and tried to solve it like the above(telescoping series). For example, $ (\sqrt i +1)^2-(\sqrt i)^2 = 2\sqrt i +1$. But it didn't work so well........ Does anyone know how to do it? My math knowledge is only within Calculus.
- Is there a generalizing method? That is, $$\sum_{i=1}^n i^x$$ where $x$ is rational(or real)