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I want to show the following:

If $\sum a_n, \sum a_n^2, \dots , \sum a_n^{k-1}$ and $\sum |a_n|^k$ are all convergent than $\prod (1+a_n)$ is also convergent.

I found this statement: $\prod (1+a_n)$ converges iff $ \sum a_n$ converges

My problem is from the Appendix of "Special Functions" of Andrews.

I am confused by the problem since I think that the statement under the link is not true. Shouldnt it be that $\sum a_n$ has to be absolutely convergent?

Back to my problem: I have no idea why he has to climd to $k$ and thant take the absolute value in order to get convergence of the product. Any help?

Averroes2
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  • If you read carefully the post in the link you provided, you will see that the question is actually "Under which extra conditions is that statement true"? Don't just read the title, read the posts. – N. S. Oct 01 '20 at 15:20

1 Answers1

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Hint

First, by ignoring eventually the first few terms you can assume WLOG that $a_n \in (-1,1)$.

$$\ln (\prod_{n=1}^N (1+a_n))=\sum_{n=1}^N \ln(1+a_n) $$

Now, approximate $\ln(1+x)$ by its $k-1$ Taylor polynomial, and show that the error in the approximation is bounded by $x^k$.

N. S.
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  • Can you show me the approximation of the $k-1$ Taylor polynomial? Is it $\frac{\log(1+a_n)-a_n+\frac{a_n^2}{2}-\frac{a_n^3}{3}+\dots+(-1)^k\frac{a_n^{k-1}}{k-1}}{a_n^k}=\frac{(-1)^{k+1}}{k}$? – Averroes2 Oct 01 '20 at 16:12
  • @Averroes2 See for example here https://math.stackexchange.com/questions/383659/finding-the-error-of-the-taylor-expansion-of-log1-x – N. S. Oct 01 '20 at 17:45
  • @Averroes2 Note that the approximation is not equality, it is LESS than the error. $$\left| \log(1+a_n) - P_{k-1}(a_n) \right| \leq |a_n|^k$$ by (4) in the link above – N. S. Oct 01 '20 at 17:48