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I have a feeling I am making a very silly mistake here but is the limit for $\lim_{n \to \infty} |\frac {zn^n}{(n+1)^{n+1}}| =|z|$ ?

just need a confirmation.

Thanks guys.

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Seems to work like this: $$ \lim_{n\to\infty} \bigg|\frac{zn^n}{(n+1)^{n+1}}\bigg|=\lvert z\rvert\lim_{n\to\infty}\bigg|\frac{n^n}{(n+1)^{n+1}}\bigg|=\lvert z\rvert\lim_{n\to\infty}\bigg|\frac{1}{n}\bigg|=0.$$ The third equality follows from expanding the denominator. We can see that the dominating term is the term with power $(n+1)$: that is $n^{n+1}$. If we divide the numerator and denominator by $n^n$ and take the limit, we get the result.

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  • $\begingroup$ Or observe that $n^n/(n+1)^{n+1}=[(n/(n+1))^n]/[n+1]<1/[n+1].$.........+1 $\endgroup$ – DanielWainfleet Feb 3 '17 at 3:20
  • $\begingroup$ That works too! $\endgroup$ – Antonios-Alexandros Robotis Feb 3 '17 at 3:21
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You should know that $\lim\limits_{n\to\infty} \left( \dfrac n {n+1}\right)^n = \dfrac 1 e,$ and what you've got is $\lim\limits_{n\to\infty} \left( \dfrac n {n+1} \right)^n \cdot \dfrac 1 {n+1}. $

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