A threshold for an exponential sum

Question

I came across a sum where I have to find the smallest $n$ so that

$$\sum_{x = 0}^n \frac{250^x}{x!} \ge \frac{e^{250}}{2}$$

I wrote a Java code and the result was 55 but with Desmos it was over 129 (I don't know exactly how much because Desmos gives "undefined value" over 129). How would you proceed in such a case? I also took the natural log of both sides because it is monotony-preserving but we have the same problem.

Answer 1

In fact, your inequality can be given a probabilistic interpretation : written under the form

$$\sum_{x = 0}^n e^{-250} \frac{250^x}{x!} \ge \frac{1}{2},$$

the first $n$ for which this inequality is true has a name : it is the median of the Poisson distribution with parameter $\lambda=250$.

In this reference, we have a "bracketing result" saying that the median $\nu$ is such that :

$$\lambda - \ln(2) \le \nu \le \lambda+\tfrac13$$

giving here $\nu=\lambda=250.$

Confirmation by this SAGE program (SAGE language supports multiprecision), where $S$, once the for-loop has been completed, is the sum on the LHS of your initial inequality :

 S=0
 c=1 ; # increment
 n=250
 for k in range(n) :
    S=S+c
    c=250*c/(k+1)
 show(S.n(30))
 t=exp(250)/2
 show(t.n(30))

Taking $n=250$ is

$$S \approx 1.8417173 \times 10^{108} < t \approx 1.8732273 \times 10^{108}$$

whereas, for $n=251$, it is :

$$S \approx 1.9362138\times 10^{108} > t \approx 1.8732273 \times 10^{108}$$

Answer 2

Using algebra and making the problem more general, consider that you look for the zero of function $$f(n)=\sum_{x = 0}^n \frac{a^x}{x!} - \frac{e^a}{2}$$ that is to say $$f(n)=\frac{e^a \,\,\Gamma (n+1,a)}{n!}-\frac{e^a}{2}$$ or, better, use Desmos to find the zero of function $$g(n)=\log \left(\frac{2 \,\Gamma(n+1,a)}{n!}\right)$$ considerind that $n$ is a continuous variable. If you make a plot, you will see that $g(n)$ is close to linearity around $n=a$.

Using asymptotics (correction made after @Gary's comment) $$\large\color{blue} {n \sim a - \frac 2 3-\frac 8{405\, a}}$$

For $a=250$, some results $$\left( \begin{array}{cccc} k & \text{Newton} & \text{Halley} & \text{Householder} \\ 0 & 250.0000000 & 250.0000000 & 250.0000000 \\ 1 & 249.3214655 & 249.3333918 & 249.3332533 \\ 2 & 249.3332507 & 249.3332543 & 249.3332543 \\ 3 & 249.3332543 & 249.3332543 & 249.3332543 \\ \end{array} \right)$$