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If $n$ is large enough, which is greater:

$(n+1) ^{n+1}$ or $(kn)^{n}$ where $k$ is a natural number?

I've plotted a graph which suggests that the second is larger, but surely the larger power should dominate in the end?

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    $\begingroup$ Divide both by $n^n$. $\endgroup$ – Daniel Fischer Nov 11 '14 at 20:17
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    $\begingroup$ Note that $(kn)^n=n^{n(log_n(k)+1)}$ so for $k>1$, $(kn)^n$ "has the larger power". $\endgroup$ – Solomonoff's Secret Nov 11 '14 at 20:53
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dividing both sides by $n^{n+1}$ gives

$\left(1+{1\over n}\right)^{n+1}$ and ${k^n\over n}$

As $n\to\infty$ the first quantity converges to $e$, but the second goes off to infinity unless $k=1$, in which case it is already trivial that $(n+1)^{n+1}$ is the larger of the two.

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previous clues are good enough for you.

Nevertheless, if you want to be more formal, you can use Newton's binomial expansion for the power $n + 1$, in which case you can compare the first one to $(kn)^n = k^n \times n^n \ldots$

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  • $\begingroup$ I don't see how comparing the expansion of the left hand side helps. $\endgroup$ – Jam Nov 11 '14 at 21:07
  • $\begingroup$ Look at the first two terms of the binomial expansion: $(n+1)^{n+1}=n^{n+1}+(n+1)n^n+$[additional positive terms]. How does this compare to $k^n n^n$? $\endgroup$ – Steve Kass Nov 12 '14 at 2:56

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