Home » Blog » Find all z such that the series (-1)n / (2n+1) • ((1-z) / (1+z))n converges

Find all z such that the series (-1)n / (2n+1) • ((1-z) / (1+z))n converges

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1} \left( \frac{1-z}{1+z} \right)^n\]

converges.


Let

    \[ b_n = \frac{1}{2n-1}, \qquad a_n = \left( \frac{1-z}{1+z} \right)^n. \]

Then the series converges if \sum a_n converges. But, since \sum a_n is a geometric series, it converges if

    \[ \left| \frac{1-z}{1+z} \right| < 1 \quad \implies \quad |1-z| < |1+z|. \]

If we write z = x + iy this inequality holds if and only if x > 0.
On the other hand,

    \[ |1-z| > |1+z| \quad \implies \quad \frac{1-z}{1+z} > 1 \]

and so

    \[ \lim_{n \to \infty} \frac{1}{2n-1} \left( \frac{1-z}{1+z} \right)^n \neq 0 \]

hence, the series diverges.

Therefore, the series converges for all z = x + iy with x \geq 0.

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