Limit is infinite or finite?

The given function is $${{Log\ z}\over {z-1}}=1- {1\over 2}(z-1)+ {1\over 3} (z-1)^2-{1\over 4}(z-1)^3+....$$ Then it is said that the function tends to $+\infty$ as $z$ tends to $0$ . But how $?$ As $z$ tends to $0$ the function tends to $$1-{1\over 2}+{1\over 3}-{1\over 4} +....$$ which is a convergent series hence the sum is $\lt \infty$ , i.e. some finite number . So how can it " tend to $+\infty$" $?$

• If you plug $z=0$ in the RHS, you get $$1+{1\over 2}+{1\over 3}+{1\over 4} +\ldots=+\infty.$$ – Did Oct 19 '15 at 6:43
You've got the wrong sign. At $z=0$ the series becomes $$1+{1\over 2}+{1\over 3}+{1\over 4} +\ldots$$ which is divergent.