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Define the integral $I_n$ as $$I_n = \int_0^1 \frac{d^n}{dx^n} \frac{(x-x^2)^n}{n!} e^xdx = a_ne+b_n $$ where $a_n$ and $b_n$ are integers. We can integrate $I_n$ by parts $n$ times $$I_n = (-1)^n \int_0^1\frac{(x-x^2)^n}{n!} \frac{d^n}{dx^n}e^xdx\\ = (-1)^n \int_0^1\frac{(x-x^2)^n}{n!} e^xdx$$ thus given an upper bound for it $$|I_n| \leq \int_0^1\frac{[\text{max}(x-x^2)]^n}{n!} e^xdx=\frac{e-1}{n!4^n}.$$

Therefore $$ |I_n| = |a_ne+b_n| \leq \frac{e-1}{n!4^n}.$$ My question is: if we replace $a_ne+b_n$ for $a_nS_n+b_n$ where $$S_n = \sum_{k=0}^n \frac{1}{k!} $$ is the upper bound for $a_nS_n+b_n$ still is $(e-1)/n!4^n?$ If not, what is the new upper bound?

edit: If I remember right $e-S_n\leq\frac{1}{n!n}$

Pinteco
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1 Answers1

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Consider $|a_n S_n + b_n|$ with $I_n = a_n e + b_n$.

Using $|a_n e +b_n |\leq \frac{e-1}{n! 4^n}$,

$|a_n S_n + b_n | \leq |a_n S_n - a_n e + a_n e + b_n| \leq |a_n S_n - a_n e |+ |a_n e + b_n| \leq \sup (a_n) \frac{1}{n!} + \frac{e-1}{n! 4^n}$.

Potentially more analysis to do on the estimation of $\sup (a_n)$

fGDu94
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