I am trying to verify the following statement mentioned in an wikipedia article (https://en.wikipedia.org/wiki/Stirling%27s_approximation#Derivation), the statement is used to derive the stirling's approximation . I also took some help from another post (Property of the remainder term in the Euler-Maclaurin formula for $\sum_{i=1}^n\log i$.). $$R_{m,n}=\lim_{n\to\infty} R_{m,n}+O(n^{-m})$$ Here's what i tried. $$R_{m,n}=\int_1^{n}\frac{P_m(x)}{mx^m}~ dx \text{ , where $P_m(x)$ is a periodic Bernoulli polynomial.}$$ \begin{align} \left|\lim_{n\to\infty} R_{m,n}-R_{m,n}\right|&=\left|\int_n^{\infty}\frac{P_m(x)}{mx^m}~ dx\right|\\ &\le\int_n^{\infty}\frac{|P_m(x)|}{mx^m}~ dx\\ \end{align} Since R.H.S is an improper integral we have
\begin{align} \lim_{t\to\infty}\int_n^{t}\frac{|P_m(x)|}{mx^m}~ dx\\ \end{align} Integrating by parts and using the property of periodic bernoulli function $P'_m({x}) = mP_{m-1}(x) $ for $m \ge 3$ we get
\begin{align} \lim_{t\to\infty}\int_n^{t}\frac{|P_m(x)|}{mx^m}~ dx = \lim_{t\to\infty}\left(\left|-\frac{P_m(x)}{m(m-1)x^{m-1}} - \frac{P_{m-1}(x)}{(m-1)(m-2)x^{m-2}} - \frac{P_{m-2}(x)}{(m-2)(m-3)x^{m-3}} ... -\frac{P_{2}(x)}{x}\right|^t_n\right)\\ \end{align}
Then evaluating integral from $t $ to $n$ we get \begin{align} \lim_{t\to\infty}F(t) = \lim_{t\to\infty}\left(\left(-\frac{P_m(t)}{m(m-1)t^{m-1}} - \frac{P_{m-1}(t)}{(m-1)(m-2)t^{m-2}} - \frac{P_{m-2}(t)}{(m-2)(m-3)t^{m-3}} ... -\frac{P_{2}(t)}{t}\right)- \left(-\frac{P_m(n)}{m(m-1)n^{m-1}} - \frac{P_{m-1}(n)}{(m-1)(m-2)n^{m-2}} - \frac{P_{m-2}(n)}{(m-2)(m-3)n^{m-3}} ... -\frac{P_{2}(n)}{n}\right)\right)\\ \end{align} \begin{align} \lim_{t\to\infty}F(t) = \lim_{t\to\infty}\left(\left(-\frac{B_m}{m(m-1)t^{m-1}} - \frac{B_{m-1}}{(m-1)(m-2)t^{m-2}} - \frac{B_{m-2}}{(m-2)(m-3)t^{m-3}} ... -\frac{B_{2}}{t}\right)- \left(-\frac{B_m}{m(m-1)n^{m-1}} - \frac{B_{m-1}}{(m-1)(m-2)n^{m-2}} - \frac{B_{m-2}}{(m-2)(m-3)n^{m-3}} ... -\frac{B_{2}}{n}\right)\right)\\ \end{align}
Finally taking the limit \begin{align} =\left(\left(0+ 0 + 0 ....+0\right)+ \frac{B_m}{m(m-1)n^{m-1}} + \frac{B_{m-1}}{(m-1)(m-2)n^{m-2}} + \frac{B_{m-2}}{(m-2)(m-3)n^{m-3}} ... +\frac{B_{2}}{n}\right)\\ \end{align} Therefore i get $O({n^{1-m}})$ , but in wikipedia it is given that it is bounded by $O({n^{-m}})$. I think i am doing something wrong here, any help is greatly appreciated -:)
Note that $$ R_{m,n} = \frac{{( - 1)^{m + 1} B_{m + 1} }}{{m(m + 1)}}\left( {\frac{1}{{n^m }} - 1} \right) + R_{m + 1,n} \\ = \frac{{( - 1)^{m + 1} B_{m + 1} }}{{m(m + 1)}}\left( {\frac{1}{{n^m }} - 1} \right) + \int_1^n {\frac{{P_{m + 1} (x)}}{{(m + 1)x^{m + 1} }}dx} , $$ whence $$ R_{m, \infty } - R_{m,n} = \frac{{( - 1)^m B_{m + 1} }}{{m(m + 1)}}\frac{1}{{n^m }} + \int_n^{ + \infty } {\frac{{P_{m + 1} (x)}}{{(m + 1)x^{m + 1} }}dx} \\ = \frac{{( - 1)^m B_{m + 1} }}{{m(m + 1)}}\frac{1}{{n^m }} + \mathcal{O}\!\left( {\frac{1}{{n^m }}} \right) = \mathcal{O}\!\left( {\frac{1}{{n^m }}} \right). $$