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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.

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    $\begingroup$ If you plug $z=0$ in the RHS, you get $$1+{1\over 2}+{1\over 3}+{1\over 4} +\ldots=+\infty.$$ $\endgroup$ – Did Oct 19 '15 at 6:43
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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.

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