Calculate:&nbsp;limn→∞e-n∑k=0nnkk!

Question

Calculate:&amp;nbsp;limn→∞e-n∑k=0nnkk!

Terry Ray · Accepted Answer

limn→∞[e−n∑k=0nnkk!]  =limn→∞[e−n[∑k=0nexp⁡(kln⁡(n)−ln⁡(k!))]  =limn→∞[e−n∑k=0nexp⁡(nln⁡(n)−ln⁡(n!)−12n[k−n]2)]  =limn→∞[e−nnnn!∑k=0nexp⁡(−12n[k−n]2)]  =12limn→∞[2πnn+12e−nn!]=12

Lakisha Archer · Accepted Answer

The sum is related to the partial exponential sum, and thus to the incomplete gamma function,

e^{- n} \sum_{k = 0}^{n} \frac{n^{k}}{k!} = e^{- n} e^{n} (n)

= \frac{Γ (n + 1, n)}{Γ (n + 1)}

since

e_{n} (x) = \sum_{k = 0} \frac{x^{k}}{k} \neq e^{x} Γ \frac{n + 1, x}{Γ} (n + 1)

. But

Γ (n + 1, n) = \sqrt{2 π} n^{n + \frac{1}{2}} e^{- n} (\frac{1}{2} + \frac{1}{3} \sqrt{\frac{2}{n π}} + O (\frac{1}{n}))

The first term in the asymptotic expansion for

Γ (n + 1, n) = \int_{n}^{\infty} d t t^{n} e^{- t}

The higher order terms are in principle straightforward to compute. Using Stirlings

alenahelenash · Accepted Answer

On this page there is a nice collection of evidence. I add another proof which also uses the Stirling formula. e−n∑k=0nnkk!=e−n∑k=0nkk(n−k)n−kk!(n−k)! limn→∞e−n∑k=1n−1eken−k2πk(1+O(1/k))2π(n−k)(1+O(1/(n−k))) limn→∞12π1n∑k=1n−11kn(1−kn)=12π∫01dxx(1−x)=Γ(12)22πΓ(1)=12

Calculate: lim_{n -> oo) e^(-n) sum_(k=0)^n (n^k)/(k!)

Answered question

Answer & Explanation

New Questions in Calculus 2