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 $" $?$
Please explain.
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.