# Domain of convergence [closed]

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.

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

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

• Could you clarify more for me? I still didn't get it.. Thanks Apr 26 '20 at 22:00
• Within parentheses find a common denominator
– zhw.
Apr 26 '20 at 22:27
• Ok.. then we get $\sum_{n=0}^{\infty}\frac{(-1)^n}{z+n}=\sum_{n=0}^{\infty}\frac{1}{(z+n)(z+(n+1))}$ Apr 26 '20 at 22:49
• $\sum_{n=0}^{\infty}|\frac{1}{(z+n)(z+(n+1))}|<\sum_{n=0}^{\infty}\frac{1}{n(n+1)}<\sum_{n=0}^{\infty}\frac{1}{n^{2}}$ under the assumption of $z$ Apr 26 '20 at 22:53
• No we're dealing with complex numbers. $<$ doesn't make much sense.
– zhw.
Apr 26 '20 at 22:54

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.

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