I have a real-valued function $f(x)=\sum_{n=x+1}^\infty \frac{1}{n^2}$ where $x \in \mathbb{N}$. I want to understand how $f(x)$ scales with respect to $x$.
One thing I tried is as follows: Since it is not hard to see that $f(x) < \frac{1}{x}$ for all $x$, using the Big-O notation, $f(x) = O\big(\frac{1}{x}\big)$. But when I actually plot the function and use computer program to find coefficients $a,b,c$ of $f(x) = \frac{a}{x} + \frac{b}{x^2} + c$ to fit the curve, it actually gives non-zero values (that are significantly greater than $0$) for both $a$ and $b$ (and $c = 0$), indicating that my first bound is not too tight. So, how can one see a tighter scaling bound of $f(x)$?
The general method is to use the Euler-Maclaurin formula. In your case, you can notice that: $$ \frac1{n^2}\sim\frac1{n}-\frac1{n+1} $$ so: $$ \sum_{n=N}^\infty\frac1{n^2}\sim \sum_{n=N}^\infty\frac1{n}-\frac1{n+1}\sim\frac1N $$ You can then improve incrementally the asymptotic expansion by killing the next leading order term with the appropriate telescoping power series.
For example, since: $$ \frac1{n^2}-\frac1{n}+\frac1{n+1}\sim \frac{1}{n^3}\sim \frac1{2n^2}-\frac1{2(n+1)^2} $$ you get: $$ \sum_{n=N}^\infty\frac1{n^2}-\frac1N\sim \sum_{n=N}^\infty\frac1{2n^2}-\frac1{2(n+1)^2}\sim\frac1{2N^2}\\ \sum_{n=N}^\infty\frac1{n^2}=\frac1N+\frac1{2N^2}+o(N^{-2}) $$
Hope this case.