Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 104471 by Dwaipayan Shikari last updated on 21/Jul/20

Σ_(n=1) ^∞ ((e^n n!)/n^n )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{n}} {n}!}{{n}^{{n}} } \\ $$

Answered by MAB last updated on 21/Jul/20

let U_n =((n!)/n^n )e^n   by stirling formula: n!∽(√(2πn))((n/e))^n   we deduce: U_n ∽(√(2πn))  hence Σ_(n=1) ^∞ ((e^n n!)/n^n ) diverges

$${let}\:{U}_{{n}} =\frac{{n}!}{{n}^{{n}} }{e}^{{n}} \\ $$$${by}\:{stirling}\:{formula}:\:{n}!\backsim\sqrt{\mathrm{2}\pi{n}}\left(\frac{{n}}{{e}}\right)^{{n}} \\ $$$${we}\:{deduce}:\:{U}_{{n}} \backsim\sqrt{\mathrm{2}\pi{n}} \\ $$$${hence}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{n}} {n}!}{{n}^{{n}} }\:{diverges} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com