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What is the domain of convergence of the series $\sum_{n=0}^{\infty}\frac{(-1)^n}{z+n}$ ?

I'm not sure how to find the domain of convergence...?

Thanks.

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Hint: For $z\notin \{0,-1,-2,\dots\},$ think about grouping the terms as follows:

$$\left (\frac{1}{z}- \frac{1}{z+1}\right ) +\left ( \frac{1}{z+2}-\frac{1}{z+3}\right )+\cdots.$$

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  • $\begingroup$ Could you clarify more for me? I still didn't get it.. Thanks $\endgroup$
    – Math1
    Apr 26 '20 at 22:00
  • $\begingroup$ Within parentheses find a common denominator $\endgroup$
    – zhw.
    Apr 26 '20 at 22:27
  • $\begingroup$ Ok.. then we get $\sum_{n=0}^{\infty}\frac{(-1)^n}{z+n}=\sum_{n=0}^{\infty}\frac{1}{(z+n)(z+(n+1))}$ $\endgroup$
    – Math1
    Apr 26 '20 at 22:49
  • $\begingroup$ $\sum_{n=0}^{\infty}|\frac{1}{(z+n)(z+(n+1))}|<\sum_{n=0}^{\infty}\frac{1}{n(n+1)}<\sum_{n=0}^{\infty}\frac{1}{n^{2}}$ under the assumption of $z$ $\endgroup$
    – Math1
    Apr 26 '20 at 22:53
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    $\begingroup$ No we're dealing with complex numbers. $<$ doesn't make much sense. $\endgroup$
    – zhw.
    Apr 26 '20 at 22:54
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So first, we would like that the terms be bounded, so it is safe to say that $-z\notin\mathbb{N}$. Now, try using the alternating series test after breaking the sum into real and imaginary parts.

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  • $\begingroup$ Isn't the AST for series of real numbers? $\endgroup$
    – zhw.
    Apr 26 '20 at 18:46
  • $\begingroup$ You can break the series into real and imaginary parts. $\endgroup$
    – Anz
    Apr 26 '20 at 18:51
  • $\begingroup$ True, but worth mentioning I think. $\endgroup$
    – zhw.
    Apr 26 '20 at 19:04
  • $\begingroup$ @zhw. Edited. Thanks. $\endgroup$
    – Anz
    Apr 26 '20 at 19:25

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