Thanks, using specfn is an improvement as sum(1/x^2,x,1,infinity); then works!
However these still don't :)
sum(1/(x+1)^2,x,1,infinity);
sum(1/x^2,x,2,infinity);
thanks,
Mike
On 30 July 2012 15:40, Arthur Norman <acn1@...> wrote:
> On Mon, 30 Jul 2012, Michael Borcherds wrote:
>
>> Hi,
>>
>> Is there any way to get Reduce to calculate this?
>> Sum(1/x^2, x, 1, infinity);
>>
>> thanks!
>>
>> Mike
>>
>
> You may not like this answer, but consider
>
> load_package specfn;
> zeta(2);
> 2
> pi
> 
> 6
>
>
>
> Arthur
>
