Is it possible? Well, yes. And, sometimes, no.
But you can always use Mathematica. What else can I say?
When symsum returns a result that is the same as what you passed in, this is its way of telling you that no solution could be found, at least not by that tool. symsum does not always offer the ability to tell you a series is divergent, as opposed to merely not convergent.
In some cases of course, symsum has no problem. As you see here, symsum can at least some of the time return the result you want.
syms n
symsum(1./n,1,inf)
That last sum is unbounded, greater than any finite number, so an INF. But next, symsum is smart enough to know the difference, for a classically problematic series.
symsum(sin(n),0,inf)
Yes, that series is divergent, and MATLAB knew the sum is not unbounded, but that no value can be applied to the sum. So it returned NaN. to indicate the difference between an unbounded sum and one which can be only represented as NaN. But then we have this sum.
symsum(n,1,inf)
DRAT. I could have sworn that Ramanujan proved the latter sum to be -1/12? Maybe I should send Ramanujan a letter pointing out that symsum disagrees? ;-) I would send it to G.H. Hardy of course, who might be able to forward it.
Anyway, that symsum fails to tell you a series was divergent is not always conclusive. It is only conclusive that symsum did not know the answer.
symsum(log(n+1)./n,1,inf)
Computers are not omniscient. (Except when they play chess against me. Then they just laugh hysterically.)
