Use Riemann-Stieltjes Integral to derive $$\sum_{n\leq N}\frac{1}{n^\alpha} = \int_1^N\frac{1}{x^\alpha}d[x]+1.$$ Using this, show that if $\alpha>1$, the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges.
I was able to understand the derivation above, except for the first equality (especially as to where the $1$ came from).
For the second equality, I am confused as to how the terms changed, I know about the reduction of a Stieltjes integral to a Riemann integral, but that can only be used when we are integrating with respect to a continuous function for which $[x]$ is not (this is the floor function, which has a discontinuity at every integer). Could anyone elaborate on these issues?
Apostol's book has the following result: if $f \in R(\alpha)$ on $[a,b]$ then $\alpha \in R(f)$ and $\int_a^{b}f d\alpha +\int_a ^{b} \alpha df=f(b)\alpha(b)-f(a)\alpha(a)$. This theorem can be used if either $f$ or $\alpha$ is continuous and the other one is of bounded variation; no need for both to be continuous.