I get $N=3$; that is, the first 4 terms in the sum are accurate to within that error.

You are watching: E^(-1/4)

To elaborate:

$$e^{0.25} approx 1.28403$$

$$sum_{k=0}^{3} frac{1}{4^k,k!} approx 1.28385$$

$$ ext{error} approx 0.000171$$

let $nin hedailysplash.tvbb N_{geq0}$, so by Maclaurin expansion there”s $ hetain(0,frac{1}{4})$ such that

$$e^{frac{1}{4}}=1+frac{1}{4}+frac{1}{4^2.2!}+ldots+frac{1}{4^n.n!}+frac{1}{4^{n+1}(n+1)!}e^{frac{1}{4} heta}$$we look for $n$ such that the remainder$$R_n=frac{1}{4^{n+1}(n+1)!}e^{frac{1}{4} heta}leq frac{3}{4^{n+1}(n+1)!}leq 10^{-3}$$we find numerically$$R_2approx 7.8 imes 10^{-3}quad ext{and}quad R_3approx 4.8 imes 10^{-4}$$so we choose $n=3$ and we have$$e^{frac{1}{4}}approx 1+frac{1}{4}+frac{1}{4^2.2!}+frac{1}{4^3.3!}approx1.2839$$

Just substitute 1/4 for the value of x in each term.By the time you get to the 5th term the value is 0.000163 which is much smaller than your tolerance. Subsequent terms, with the factorial on the bottom, will get much smaller very rapidly.Their total will not affect the first 3 decimal places of the answer.Essentially you only need 4 terms of the series.

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