What is a good approximation for the function:
$$S_{N,k} = \sum_{i=N}^\infty {\frac{1}{i^k}}$$
when $k$ is a given constant (2, 3 or 4) and $N$ is large?
$S_{N,k}$ is a decreasing function of $N$; I am interested to know how fast it decreases. For example, is it true that $S_{N,k} = O(1/N)$?
On
One can approximate $\int_N^\infty x^{-k}dx$ in two simple ways: The sum of rectangles approximation will be a bit low, and is given by
$$S_{N,k} = N^{-k}+ \sum_{i=N+1}^\infty {\frac{1}{i^k}} < N^{-k}+ \int_N^\infty x^{-k}dx = N^{-k}+\frac{1}{k-1}N^{1-k} $$
The trapezoid rule approach is a bit high (since the function is concave upward), and is given by $$S_{N,k} =\frac12 N^{-k} + \left( \frac12 N^{-k} + \sum_{i=N+1}^\infty {\frac{1}{i^k}} \right) > \frac12 N^{-k}+ \int_N^\infty x^{-k}dx = \frac12 N^{-k}+ \frac{1}{k-1}N^{1-k} $$ Combining these, $$ \frac12 N^{-k}+ \frac{1}{k-1}N^{1-k} < S_{N,k} < N^{-k}+\frac{1}{k-1}N^{1-k} $$
Since for large $N$ and fixed $k$ the $\frac{1}{k-1}N^{1-k}$ terms are much larger than the $N^{-k}$ terms, $$ S_{N,k} = O(N^{1-k} = O \left( \frac{1}{N^{k-1}}\right)$$
So for $k \geq 2$ it is also true that $$ S_{N,k} = O\left( \frac{1}{N} \right) $$ but for $k>2$ the above stronger form is a "better" approximation.
You can approximate the sum using an integral. If $j < x < j+1$ you have $\frac{1}{j^k} > \int_j^{j+1} \frac{1}{x^k} dx > \frac{1}{(j+1)^k}$. This will allow you to develop both an upper and a lower bound on your sum and you should find that $S_{N,k} = O(1/N^{k-1})$.