Question 696891

{{{an = n^2/n!}}} for n ≥ 0.

{{{n=0}}}
{{{an = 0^2/0!=0/0=0}}}

{{{n=1}}}
{{{an = 1^2/1!=1/1=1}}}

{{{n=2}}}
{{{an = 2^2/2!=4/(2*1)=4/2=2}}}

{{{n=3}}}
{{{an = 3^2/3!=9/(3*2*1)=9/6=3/2}}}

{{{n=4}}}
{{{an = 4^2/4!=16/(4*3*2*1)=16/24=2/3}}}


so, the first five terms of the sequence defined by {{{an = n^2/n!}}} are:

{{{0}}},{{{1}}},{{{2}}},{{{3/2}}},{{{2/3}}}