To show that the sequence converge, we need to show that the limit of the nth term when n tends to infinity is a constant.

Usually we can not find limits of factorials, so we need to rewrite our problem a little differently.

`(10^n)/(n!)=(10^10)/(10!)xx(10/11)xx(10/12)xx(10/13)xx...(10/n)` From that we can conclude:

`0<(10^n)/(n!)<(10^10/10!)*(10/11)^(n-10)`

If we can show that the left hand side and right hand side of the above inequality have the same limit, then the middle expression will have the same limit.

Note that `lim_(n->oo)0=0`

and since 10/11 is less than one we have `lim_(n->oo)(10^10/(10!))xx(10/11)^(n-10)=(10^10)/(10!)xx0=0`

Since the previous two limits equal the same number, then

`lim_(n->oo)(10^n)/(10!)=0`

**Hence your sequence converge.**