How to write formally within the context of mathematics that: "given two series S1 and S2 and they are subtracted each other coming from a proved identity that is true and the result of this subtraction is a known finite number (real number) (which is valid) the two series S1 and S2 are convergent necessarily because the difference could not be divergent as it would contradict the result of convergence? I need that definition within a pure mathematical scenario ( I am engineer).
" Given S1- S2 = c , if c is a finite and real number, and the expression S1-S2 = c comes from a valid deduction, then, S1 and S2 are both convergent as mandatory."
Best regards
Carlos
Cite:
https://www.quora.com/If-the-difference-of-two-series-converges-do-the-two-series
Mark Gritter
No. Consider the series ... in the above link example/ answer by Mark Gritter.
Both are divergent series, but the difference between the series is ... a convergent series.
https://qr.ae/prh7aB
Steftcho P. Dokov Yes, I have investigated and you are right. However, in my case the second series S2 is likely convergent as there are too many terms on the expansion and there is not jumps discovered. However, I will be looking for a formal analytical convergence for that second series which will give me guarantee of the total convergence I expect.Many thanks.
Carlos Hernan Lopez Zapata , adding to the first answer and in relation to your last post, it may be useful to know that if a real series Σxn converges, then the sequence (xn) of terms of the series converges to zero, but the opposite does not necessarily hold, with a counterexample being the harmonic series Σ1/n, which diverges to positive infinity in spite of the fact that (1/n) converges to 0. Hence there exist cases where a series “seems” to converge, but actually it does not.
An if and only if criterion for the convergence of a real series is the Cauchy criterion, which states that a real series converges if and only if the sequence of partial sums of the series is a Cauchy sequence. Formally, Σxn converges if and only if for every ε>0 there exists a natural number Ν such that for all natural numbers m,n, if m>n>N, then abs(sm-sn)
The matter here is the S2 which computationallu converges by resting even 10.000 terms of expansion. But it is a series without an analytical comprehension. I can’t use some expression lemma like “for large enough n.... the expansion gives us a convergent series S2...”?
If it is possible to derive an upper bound for the nth term of S2, then you can compare it with the nth term of a known series and maybe a comparison test may be applicable, or if it is possible to compare the nth term or an asymptotic expression of the nth term of S2 directly with the nth term of a known series, then a comparison test may be applicable. Similarly, if it is possible to derive an expression of the ratio of the nth term of S2 to the nth term of a known series, then a ratio test may be applicable. There exist many convergence tests for series; some may be applicable in your case.
- Badges
- Science topic