Editing Radius of convergence (section)

===A more complicated example===

Consider this power series:

:<math>\frac z {e^z-1}=\sum_{n=0}^\infty \frac{B_n}{n!} z^n </math>

where the rational numbers ''B''<sub>''n''</sub> are the [[Bernoulli numbers]].  It may be cumbersome to try to apply the ratio test to find the radius of convergence of this series.  But the theorem of complex analysis stated above  quickly solves the problem.  At ''z'' = 0, there is in effect no singularity since [[removable singularity|the singularity is removable]].  The only non-removable singularities are therefore located at the ''other'' points where the denominator is zero.  We solve

:<math>e^z - 1 = 0</math>

by recalling that if {{math|1=''z'' = ''x'' + ''iy''}} and {{math|1=''e''{{i sup|''iy''}} = cos(''y'') + ''i'' sin(''y'')}} then

:<math>e^z = e^x e^{iy} = e^x(\cos(y)+i\sin(y)),</math>

and then take ''x'' and ''y'' to be real.  Since ''y'' is real, the absolute value of {{math|cos(''y'') + ''i'' sin(''y'')}} is necessarily 1.  Therefore, the absolute value of ''e''{{i sup|''z''}} can be 1 only if ''e''{{i sup|''x''}} is 1; since ''x'' is real, that happens only if ''x'' = 0.  Therefore ''z'' is purely imaginary and {{math|1=cos(''y'') + ''i'' sin(''y'') = 1}}.  Since ''y'' is real, that happens only if cos(''y'') = 1 and sin(''y'') = 0, so that ''y'' is an integer multiple of&nbsp;2{{pi}}.  Consequently the singular points of this function occur at

: ''z'' = a nonzero integer multiple of&nbsp;2{{pi}}''i''.

The singularities nearest 0, which is the center of the power series expansion, are at ±2{{pi}}''i''.  The distance from the center to either of those points is 2{{pi}}, so the radius of convergence is&nbsp;2{{pi}}.