TITLE: Maclaurin of $f(x)=\ln\left(\sum_{k=0}^{1000}\frac{x^k}{k!}\right)$
QUESTION [5 upvotes]: The 1001st order is required.
Here is how I went about the question:
We know that near $x=0$:
$$e^x=\sum_{k=0}^{1000}\frac{x^k}{k!}+\frac{x^{1001}}{1001!}+o(x^{1001})$$
So,
$$e^x-\frac{x^{1001}}{1001!}+o(x^{1001})=\sum_{k=0}^{1000}\frac{x^k}{k!}$$
And after taking the logarithm of both sides,
$$f(x)=\ln\left(e^x-\frac{x^{1001}}{1001!}+o(x^{1001})\right)$$
But what do I do next? I intuitively understand that the answer should be:
$$f(x)=x-\frac{x^{1001}}{1001!}+o(x^{1001})$$
But how do I prove it rigorously?
Thank you!

REPLY [2 votes]: Re-posting my comment as an answer since it resolved the question:

Factor out $e^x$ in the argument of $\ln$, then apply the fact that the log of a product is the sum of the logs. After this you can apply the Taylor series for the logarithm.