# Tag Info

First we can see that ${\frac{(6k+1)^{k}}{(2k+5)^{k}}}$ behaves asymptotically as $3^k$, while $(z-2i)^k$ is divergent if $\|z-2i\|>1$, however when $\|z-2i\|<1$ it is convergent. For $\|z-2i\|=1$ this criterion is not conclusive. On the other hand we can carefully extend this argument to the full sequence, so we need only \$\|z-2i\| < ...