P-series

The series ${\sum }_{n=1}^{\infty }{n}^{-p}$ converges exactly when $p>1$.

Fix an integer $N>1$ and a real exponent $p>1$, and put $S_p={\sum }_{n=N}^{\infty }{n}^{-p}$. There is a unique shift $k_p<N$ such thatIntegration evaluates the left side as $(N-k_p{)}^{1-p}/(p-1)$, yielding the displayed formula. Comparing the decreasing function $x^{-p}$ with unit rectangles provesThe integral is a strictly increasing function of its shift, so $0<k_p<1$. This is an exact reparametrization of a tail, not a claim that its shift is independent of $p$; interpolating shifts for different exponents is an additional approximation.

For $p>0$ and $n\ge 2$, integrating the decreasing derivative magnitude of $x^{-p}$ gives $(n-1{)}^{-p}-n^{-p}=p{\int }_{n-1}^n{x}^{-p-1}dx>p{n}^{-p-1}$. Put $p=k-1$ and sum this telescoping series to obtain the bound in the title, strict for $N>1$. The partial sums are increasing and bounded, proving convergence of the p-series when $k>1$. This gives a direct comparison test for series proof and a finite upper bound without evaluating the sum.