I am studying the proof of the below Lemma and have some questions regarding the bounds for truncated Taylor expansion.
We take $N$ some large number and $1\le M \le N \le t$ and $r=\frac{5.01 \log t}{\log N}$ which is just some number to get the bound. $x$ and $y$ are just some dummy variables set up so that when we take the sum $\sum_{N<n\le N+M} n^{-it}$ we have $xy/n \le N^{-1/5}$.
The question I have here is when we apply Taylor expansion to $\log(1+\frac{xy}{n})$ why do we get $$\log(1+\frac{xy}{n})=\sum_{j=1}^{\lfloor r \rfloor} \frac{(-1)^{j-1}}{j}(\frac{xy}{n})^j + O\big((\frac{xy}{n})^{5.01(\log t)/\log N}\big),$$ i.e. how can we just truncate the logarithm to the first $r$ terms and bound the rest by $(\frac{xy}{n})^{r}$? This isn't even the $\lfloor r \rfloor +1$th term of the Taylor expansion so I am not sure where this bound comes from.
Finally, the reason we do this expansion for $\log$ is actually because we apply it to $(1+xy/n)^{-it}=e^{-it \log(1+\frac{xy}{n})}.$ Because we have $xy/n \le N^{-1/5}$ we have $O\big((\frac{xy}{n})^{5.01(\log t)/\log N}\big) = O(t^{-1/500})$. And for the last part of the proof we use the fact that $\exp\{O(t^{-1/500})\} = 1 + O(t^{-1/500})$. Why can we absorb the infinite sum of the terms by the first order of the Taylor expansion?
I thought the second result could hold because if we let $a=t^{-1/500}$ then the remaining sum after the first term is $t^{-1/500} + t^{-2/500}/2! + t^{-3/500}/3!+\cdots$ but this is $\ll a+a^2+ a^3 +\cdots = a(1+a+a^2+\cdots)$ where $a<1$ and hence by geometric progression the infinite sum inside the bracket converges to $1/(1-a)$. But this is not a constant and depends on $a$ so how can I show it gets swallowed into the $O(a)$ bound? I think the bound for logarithm can be achieved similarly but then I run into the same problem.
Update I think my method works here because we assume $t$ is a very large number so we have some constant $c$ that is $1<c<t^{1/500}$ then $t^{-1500}=a<1/c<1$ hence we can bound $1+a+a^2+\cdots $ by $1+1/c+1/c^2 + \cdots = 1/(1-1/c)$ which is just a constant.
Now we can do something similar for the logarithm case as we have $xy/n \le N^{-1/5}$ so again we can treat $(xy/n)^j$ part as a geometric progression and in fact this is a worse case than the given sum because with the $1/j$ term the series would converge faster. Then similarly we should get the bound by the next term $O((\frac{xy}{n}^{\lfloor r \rfloor + 1}))$ but since $xy/n <1 $, we should have $\{xy/n\}^{\lfloor r \rfloor + 1} \le \{xy/n\}^r$ which gives the result of the Lemma.
I would greatly appreciate if someone verify my reasoning or suggest something else. I have looked at the Taylor estimate for the remainder in https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder but I don't understand how this can be applied here without a clear context on where the intervals for the function is in and the lower and upper bound for the next order derivative.
The other terms are $O(t^{-k/500})$ for $k \ge 2$ and these are swallowed up by $O(t^{-1/500})$.