~~~
The Maxima function
(%i1) f(n1,n2):=sum(1.0/(10)^k,k,n1,n2);
(%o1) f(n1,n2):=sum(1.0/10^k,k,n1,n2)
Can be used to get the numeric look and feel for the result of this sum.
For example,
(%i17) f(3,5);
(%o17) 0.00111
(%i2) f(-2,5);
(%o2) 111.11111
(%i2) f(-6,-4);
(%o2) 1110000.0
For n1 less than about -300 or n2 lager than about 16 the numerical result is outside
the usual the range and precision of the default floating point model.
By inspection, the decimal representation of the sum result number can be specified to any n1 < n2
~~~
-----Original Message-----
>
> <..> How would a new user figure that out? >
>> <..> tell me that the answer to the limit is "1"