In the book "Theory of Functions of a Complex Variable" by A. I. Markushevich, the extended definition of convergence of an infinite product is as follows: Given an infinite product $ \mathop{\prod}_{n=1}^{\infty} u_n $ with finitely many zero factors, let $ N \ge 0 $ be any integer such that all the factors $ u_{N+1} ,u_{N+2},... $ are nonzero, and let $ v $ be any integer greater than $ N $, then we say that $ \mathop{\prod}_{n=1}^{\infty} u_n $ converges if the limit $$ \lim_{v \to \infty} \mathop{\prod}_{n=1}^v u_n = \lim_{v \to \infty} \mathop{\prod}_{n=1}^N u_n \cdot \mathop{\prod}_{n=N+1}^v u_n = \mathop{\prod}_{n=1}^N u_n \cdot \lim_{v \to \infty} \mathop{\prod}_{n=N+1}^v u_n$$ exists. From this definition, the author assert that "an infinite product vanishes if and only if at least one of its factors vanishes." Is infinite product $ \mathop{\prod}_{n=1}^{\infty} \frac 1n $ a counterexample to this assertion?

  • $\begingroup$ Note: $ \mathop{\prod}_{n=1}^0 u_n $ is defined as equal to 1. $\endgroup$ – Song Wang Sep 20 '15 at 7:55
  • $\begingroup$ The transcription you provide is very strange. First, if some $u_n$ are zero, then of course the infinite product should exist and be defined as zero. Second, as you note, it is simply not true that "an infinite product vanishes if and only if at least one of its factors vanishes." For a simpler example yet, consider $u_n=\frac12$ for every $n$. $\endgroup$ – Did Sep 20 '15 at 8:00
  • $\begingroup$ @Did The transcription is copied directly from the book. In the book, an infinite product with complex nonzero factors is said to converge if the limit of the sequence of partial products converges to a nonzero limit. Otherwise, the infinite product is said to be divergent. $\endgroup$ – Song Wang Sep 20 '15 at 8:15
  • $\begingroup$ Yes this is the usual convention. Then every infinite product with at least one zero term diverges--in contradiction with the passage you quote. $\endgroup$ – Did Sep 20 '15 at 8:19
  • $\begingroup$ @Did If $ \lim_{v \to \infty} \mathop{\prod}_{n=N+1}^v u_n $ is required to converges, that is, we exclude the case that $ \lim_{v \to \infty} \mathop{\prod}_{n=N+1}^v u_n = 0 $, then the assertion seems to be right. $\endgroup$ – Song Wang Sep 20 '15 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.