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Question Show that the series $\sum _{n=1}^{\infty }\dfrac {nz^{n-1}} {z^{n}-\left( 1+\dfrac {1} {n }\right) ^{n}}$ converges absolutely if $\left| z\right| < 1$.

Answer For, when $\left| z\right| < 1$,$\left| z^{n}-\left( 1+\dfrac {1} {n}\right) ^{n}\right| \geq \left( 1+\dfrac {1} {n}\right) ^{n}-\left| z^{n}\right| \geq 1+1+\dfrac {n-1} {2n}+\ldots -1>1 $, so the moduli of the terms of the series are less than the corresponding terms of the series $\sum _{n=1}^{\infty }n\left| z^{n-1}\right| $; but this latter series is absolutely convergent, and so the given series converges absolutely.

I am having a trouble following the reason how the first 2 inequalities in the solution hold. I am sure this must be a trivial question, but i am dumb. Any help would be much appreciated.

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In the first inequality we just used $|a-b|\geq |a|-|b|$ where $a=\left(1+\frac 1n\right)^n$ and $b=z^n$, then we use the fact that $-|z|^n\geq -1$ to get, thanks to the formula which gives the expansion of $(a+b)^n$ $$\left|z^n-\left(1+\frac 1n\right)^n\right|\geq \left(1+\frac 1n\right)^n-1=1+n\frac 1n+\sum_{k=2}^n\binom nk\frac 1{n^k}-1=1+\sum_{k=2}^n\binom nk\frac 1{n^k}>1.$$

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Thanks that was very helpful i suspected similar reasons but was n't sure. – Comic Book Guy Mar 1 '12 at 19:05

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