$\sum_{1}^{\infty } (\frac{n - i}{i - 2n})^{n} (z-2)^{n} $ I tried and got $\frac{1}{2}\left | z - 2\right | < 1$

It`s correct?


The root test gives

$$\lim_\infty |a_n|^\frac 1n=\lim_\infty \left|\frac {n-i}{i-2n}\right|=$$

$$\lim_\infty \frac {\sqrt {n^2+1}}{\sqrt {1+4n^2}}=\frac {1}{2} $$

the radius of convergence is $R=2$

the series converges if $$|z-2|<2$$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.