Where is the contradiction? $\int_n^\infty {\frac{B_{2m}-B_{2m} (\left\{ t\right\})}{(2m)!}f^{(2m)}(t)dt}\sim n^{\alpha + 1} \mathcal{O}(n^{ - 2m} )$

Question

This question is why the following derivation seems to give a different answer to the result mentioned in @Gary 's answer

$$I_R=\int_n^{ + \infty } {\frac{{B_{2m} - B_{2m} (\left\{ t \right\})}}{{(2m)!}}\frac{{\alpha !}}{{(\alpha - 2m)!}}t^{\alpha - 2m} dt} = n^{\alpha + 1} \mathcal{O}(n^{ - 2m} )\tag{0}$$

where $B_{2m}$ is Bernoulli number, $B_{2m}(x)$ is Bernoulli polynomial, and $\{t\}=t-\lfloor t\rfloor$

Let $f(t)=t^\alpha$ and we have $f^{(2m)}(t)=\frac{{\alpha !}}{{(\alpha - 2m)!}}t^{\alpha - 2m}$. Let $m$ to be large enough such that $2m-1>\alpha$

$$\begin{align} I_R&=\int_n^\infty {\frac{{B_{2m} - B_{2m} (t-\lfloor t\rfloor)}}{{(2m)!}}f^{(2m)}(t)~ dt}\\ \\ I_R&=\frac{B_{2m}}{(2m)!}f^{(2m-1)}(t)|_n^\infty-\frac{1}{(2m)!}\int_n^{ \infty } B_{2m} (t-\lfloor t\rfloor) f^{(2m)}(t)~ dt\\ \\ I_R&=-\frac{B_{2m}}{(2m)!}\frac{{\alpha !}}{{(\alpha+1 - 2m)!}}~n^{\alpha+1 - 2m}-\frac{1}{(2m)!}\sum_{k=n}^\infty\int_k^{ k+1 } B_{2m} (t-k) f^{(2m)}(t)~ dt\\ \\ I_R&=-\frac{n^{\alpha+1}}{\alpha+1}\binom{\alpha+1}{2m}\frac{B_{2m}}{n^{2m}}-\frac{1}{(2m)!}\sum_{k=n}^\infty I_{k,2m} \tag{1} \end{align}$$

Where we define:

$$I_{k,p}=\int_k^{ k+1 } B_{p} (t-k) f^{(p)}(t)~ dt \tag{2}$$

Due to the special relation between Bernoulli polynomial and Bernoulli number, we start from $p=2$ to demonstrate the calculation, and deal with $p=1$ case later. Perform integration by part, we get:

$$\begin{align} I_{k,2}&=B_2(t-k)f'(t)|_k^{k+1}-\int_k^{ k+1 } B'_{2} (t-k) f'(t)~ dt\\ \\ \noindent\text{Note:}~~ B'_s(x)=sB_{s-1}(x),~~~s=1,2,3...\\ \\ I_{k,2}&=B_2(1)f'(k+1)-B_2(0)f'(k)-2\int_k^{ k+1 } B_{1} (t-k) f'(t)~ dt\\ \\ \noindent\text{Note:}~~ B_p(1)=B_p(0)=B_p,~~~p=2,3...\\ \\ I_{k,2}&=B_2\cdot\left[f'(k+1)-f'(k)\right]-2I_{k,1}\\ \\ \noindent\text{Keep going, we have}\\ \\ I_{k,3}&=B_3\cdot\left[f''(k+1)-f''(k)\right]-3I_{k,2}\\ \\ ...\\ \\ I_{k,p}&=B_p\cdot\left[f^{(p-1)}(k+1)-f^{(p-1)}(k)\right]-p\cdot I_{k,p-1}\\ \end{align}$$

Justify this recursive equation and we can get following formula: $$I_{k,p}=\frac{p!}{\alpha+1}\sum_{i=2}^p \binom{\alpha+1}{i}\cdot B_i\cdot (-1)^{p-i}\cdot\left[(k+1)^{\alpha-i+1}-k^{\alpha-i+1}\right]+(-1)^{p-1}\cdot p!\cdot I_{k,1}\tag{3}\\ $$

Now, let $p=2m$, and note $B_i=0$ for $i=3,5,7,...$

$$I_{k,2m}=\frac{(2m)!}{\alpha+1}\sum_{i=2}^{2m} \binom{\alpha+1}{i}\cdot B_i\cdot\left[(k+1)^{\alpha-i+1}-k^{\alpha-i+1}\right]-(2m)!\cdot I_{k,1}\tag{4}\\ $$

Take summation on $k$

$$\begin{align} \sum_{k=n}^\infty I_{k,2m}&=\frac{(2m)!}{\alpha+1}\sum_{i=2}^{2m} \binom{\alpha+1}{i}\cdot B_i\cdot\sum_{k=n}^\infty\left[(k+1)^{\alpha-i+1}-k^{\alpha-i+1}\right]-(2m)!\cdot \sum_{k=n}^\infty I_{k,1}\\ \\ \sum_{k=n}^\infty I_{k,2m}&=-\frac{(2m)!}{\alpha+1}\sum_{i=2}^{2m} \binom{\alpha+1}{i}\cdot B_i\cdot n^{\alpha-i+1} -(2m)!\cdot \sum_{k=n}^\infty I_{k,1}\\ \\ \sum_{k=n}^\infty I_{k,2m}&=-(2m)!~\frac{n^{\alpha+1}}{\alpha+1}\sum_{i=2}^{2m} \binom{\alpha+1}{i}\cdot \frac{B_i}{n^i} -(2m)!\cdot \sum_{k=n}^\infty I_{k,1}\tag{5} \end{align}$$

Now, let's compute $I_{k,1}$. Go back to Eq.$(2)$, perform integration by part:

$$\begin{align} I_{k,1}&=B_1(t-k)f(t)|_k^{k+1}-\int_k^{ k+1 } B_{0} (t-k) f(t)~ dt\\ \\ \noindent\text{Note:}~~ B_0(x)=B_0=1\\ \\ I_{k,1}&=B_1(1)f(k+1)-B_1(0)f(k)-B_{0}\cdot\int_k^{ k+1 } f(t)~ dt\\ \\ \noindent\text{Note:}~~ B_1(0)=B_1=-\frac{1}{2},~~~B_1(1)=-B_1=\frac{1}{2}\\ \\ I_{k,1}&=\frac{1}{2}\left[(k+1)^\alpha+k^\alpha\right]-\frac{1}{\alpha+1}\left[(k+1)^{\alpha+1}-k^{\alpha+1}\right]\\ \end{align}$$

Take the sum:

$$\begin{align} \sum_{k=n}^\infty I_{k,1}&=\frac{1}{2}\sum_{k=n}^\infty\left[(k+1)^\alpha+k^\alpha\right]-\frac{1}{\alpha+1}\sum_{k=n}^\infty\left[(k+1)^{\alpha+1}-k^{\alpha+1}\right]\\ \\ \sum_{k=n}^\infty I_{k,1}&=\sum_{k=n}^\infty k^\alpha-\frac{1}{2}n^\alpha+\frac{n^{\alpha+1}}{\alpha+1}\tag{6} \end{align}$$

Plug Eq.$(6)$ into Eq.$(5)$

$$\begin{align} \sum_{k=n}^\infty I_{k,2m}&=-(2m)!~\frac{n^{\alpha+1}}{\alpha+1}\sum_{i=2}^{2m} \binom{\alpha+1}{i}\cdot \frac{B_i}{n^i} -(2m)!\cdot \left(\sum_{k=n}^\infty k^\alpha-\frac{1}{2}n^\alpha+\frac{n^{\alpha+1}}{\alpha+1}\right)\\ \\ \sum_{k=n}^\infty I_{k,2m}&=-(2m)!~\frac{n^{\alpha+1}}{\alpha+1}\sum_{i=0}^{2m} \binom{\alpha+1}{i}\cdot \frac{B_i}{n^i} -(2m)!\cdot \sum_{k=n}^\infty k^\alpha\tag{7}\\ \end{align}$$

Plug Eq.$(7)$ into Eq.$(1)$ $$\begin{align} I_R&=-\frac{n^{\alpha+1}}{\alpha+1}\binom{\alpha+1}{2m}\frac{B_{2m}}{n^{2m}}+\frac{n^{\alpha+1}}{\alpha+1}\sum_{i=0}^{2m} \binom{\alpha+1}{i}\cdot \frac{B_i}{n^i} +\sum_{k=n}^\infty k^\alpha\\ \\ I_R&=\frac{n^{\alpha+1}}{\alpha+1}\sum_{i=0}^{2m-2} \binom{\alpha+1}{i}\cdot \frac{B_i}{n^i} +\sum_{k=n}^\infty k^\alpha \tag{8} \end{align}$$

For Eq.$(8)$

$$\begin{align} \frac{n^{\alpha+1}}{\alpha+1}\sum_{i=0}^{2m-2} \binom{\alpha+1}{i}\cdot \frac{B_i}{n^i}&\sim \frac{n^{\alpha+1}}{\alpha+1} \left(1+ \frac{c_2}{n^2}+...+ \frac{c_{2m-2}}{n^{2m-2}}\right)\\ \\ \sum_{k=n}^\infty k^\alpha &\sim \frac{n^{\alpha+1}}{\alpha+1} \end{align}$$

$$\Longrightarrow I_R \sim n^{\alpha+1} \mathcal{O}(1 )$$

But why Eq.$(0)$ gives $I_R \sim n^{\alpha + 1} \mathcal{O}(n^{ - 2m} )$ ?