# Find radius of convergence for a complex power series

Find the radius of convergence of the power series $\sum_{i=0}^\infty \frac {(-1)^k}{k}z^{k(k+1)}$. Discuss the convergence for z = 1, 1 and i.

I treated this problem as a complex power series one. I find that radius r is $\lim\limits_{x \to \infty} \lvert\frac{\frac {(-1)^k}{k}}{\frac {(-1)^{k+1}}{k+1}}\rvert=1$. But I am not sure if this is correct, since the power of z is n(n+1) instead of n.

If the radius of convergence is 1, when z=1 or -1, the series should be convergent. But we can't compare the value between 1 and i. Will that lead to an inconclusive result?

• Radius of convergence that you have calculated is correct but you should have rather used d'Alembert's root test because it will help you clarify your doubt regarding $n$ or $n(n+1)$.Furthermore,ratio test is applicable in some restricted cases(when limit of ratio exists) but d'Alembert's test and Cauchy -hadamard formula can alway be applied. – Suraj_Singh Sep 14 '15 at 8:06
• To determine the radius of convergence $R$, the simplest approach, by far, is to determine the values of $|z|$ for which the sequence $z_k=(-1)^kz^{k(k+1)}/k$ converges to zero (then $|z|\leqslant R$), respectively to infinity (then $|z|\geqslant R$). This fully determines (and with no sweat) the value of $R$. – Did Sep 14 '15 at 18:20
• "If the radius of convergence is 1, when z=1 or -1, the series should be convergent." Not necessarily. – Did Sep 14 '15 at 18:21
• If z is 1 or -1, it is always smaller or equate to R. Therefore, the series should be convergent. Isn't that correct? – Leila Sep 14 '15 at 20:21
• "If z is 1 or -1, ... the series should be convergent." Not at all. Exercise: Find a simple counterexample to this assertion. – Did Sep 15 '15 at 12:06