Improper Integrals and Series (convergence and divergence)

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When an integral with an infinite boundary converges, it yields a finite limit, indicating convergence. Conversely, a series with an upper limit of infinity converges if the limit results in a finite number. The misconception arises when interpreting the limit of an infinite series; a finite limit does not imply divergence but rather convergence. The example of a series summing to a finite value illustrates this point clearly. Additionally, divergence can occur through other means, such as oscillating series.
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Is it safe to say when an integral has an infinite boundary \int_n^∞ a_{n} and the limit yields a finite number, then the integral is said to converge.

And when a series has an upper limit of infinity \sum_n^{∞}a_{n} and the limit yields a finite number, then the series is said to diverge.
 
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If the limit of an infinite series yields a finite number, the series converges. Why would you say opposite things about series and integrals?
 
Oh, I made a conceptual mistake. I was thinking that if the limit of an infinite series yields a finite number, it meant that the terms of the series would "taper off" at some constant term. And that this constant term would be added up indefinitely, thus ∞ = divergent. For instance, if the limit of an infinite sequence yielded the finite number, 1/4, this meant:


a_{1}+ a_{2} + a_{3} + a_{4} + a_{5}...1/4 + 1/4 + 1/4 + 1/4 + 1/4... = ∞ , thus divergent


I understand now that the finite number, 1/4, in the limit of an infinite series means, that the actual sum of the series is a finite number, thus converges.

a_{1}+ a_{2} + a_{3} + a_{4} + a_{5}... = 1/4 ,thus convergent.

Thanks for the correction.
 
Note that "yielding an infinite number" is not the only way a series can diverge.
The series \sum_{n= 0}^\infty (-1)^n is also divergent.
 

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