Problem with symsum using infinity as the upper limit of the variable.

I'm not sure if I'm using symsum correctly, why can't matlab give me an aswer when m is equal to infinity?

>> m=1000;

syms n;

x=[1 1/2 1/3 1/3*2 1/4 1/5 1/6 1/7];

a=symsum(((-(1./(2.^n)))*(x.^0)).*(abs(sign((round(n*x))-n*x))-1),n,1,m);

double (a')

ans =

    1.0000
    0.3333
    0.1429
    0.1429
    0.0667
    0.0323
    0.0159
    0.0079

>> m=inf;

syms n;

x=[1 1/2 1/3 1/3*2 1/4 1/5 1/6 1/7];

a=symsum(((-(1./(2.^n)))*(x.^0)).*(abs(sign((round(n*x))-n*x))-1),n,1,m);

>> a'

ans =

             -conj(sum(1/2^n*(abs(sign(round(n) - n)) - 1), n == 1..Inf))
         -conj(sum(1/2^n*(abs(sign(round(n/2) - n/2)) - 1), n == 1..Inf))
         -conj(sum(1/2^n*(abs(sign(round(n/3) - n/3)) - 1), n == 1..Inf))
 -conj(sum(1/2^n*(abs(sign(round((2*n)/3) - (2*n)/3)) - 1), n == 1..Inf))
         -conj(sum(1/2^n*(abs(sign(round(n/4) - n/4)) - 1), n == 1..Inf))
         -conj(sum(1/2^n*(abs(sign(round(n/5) - n/5)) - 1), n == 1..Inf))
         -conj(sum(1/2^n*(abs(sign(round(n/6) - n/6)) - 1), n == 1..Inf))
         -conj(sum(1/2^n*(abs(sign(round(n/7) - n/7)) - 1), n == 1..Inf))

I'd like to find a way to get the exact values for a, which should be:

    1/1
    1/3
    1/7
    1/7
    1/15
    1/31
    1/63
    1/127

Answer 1

[Thomas]

instead of a' try

vpa(a')

You might not get all the value but close..

Answer 2

[Roger Stafford]

Tristan, you and I know the precise values of those infinite sums:

-conj(sum(1/2^n*(abs(sign(round(n) - n)) - 1), n == 1..Inf))
-conj(sum(1/2^n*(abs(sign(round(n/2) - n/2)) - 1), n == 1..Inf))
etc.

because we are intelligent human beings, but clearly in this case 'symsum' falls short of being able to do that. (I know a number of humans who would also not be able to succeed in that endeavor.) Where you gave m the finite value 1000, it must have painstakingly evaluated those expressions for each value of n to obtain an answer, but infinity proved too much for it. Perhaps one of these days Mathworks will revise 'symsum' to be smart enough to handle infinity in such a case. It is able to handle infinity in a number of other situations.