How do you justify bringing the sums into the integral? The sum is not absolutely convergent, so some care is required. Is there, for an example, 3 applications of dominated convergence that would let us take the individual sums into the integral one at a time? The value of the sum is highly dependent on the ordering (for example, you can’t reindex by grouping triples (a,b,c) based on n=a+b+c, because there are (n+2)(n+1)/2 such terms and that would give you the sum of (-1)^(n)*(n+2)(n+1))/2 *1/(n+1)=sum (-1)^n (n+2)/2, which is clearly divergent)
@ChidexOnlineMathClass01
Ай бұрын
The sum is convergence. It would have diverge if the bottom was (a+b+c). For more information, you can check on wolfram alpha
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