Me and a friend was contemplating the ratio of even to odd number in Pascals triangle. After some thought we arrived at a ratio like this when looking at a triangle of 2(adsbygoogle = window.adsbygoogle || []).push({}); ^{k-1}rows.

[itex]\frac{2^{2k-3}+2^{k-2}}{3^{k-1}} - 1 [/itex]

As expected, this ratio grows exponentially. (I.e. almost all numbers in Pascals triangle will be even) However, we then started to wonder with what base this will grow, I.e. we wondered about

[itex] \lim_{k \rightarrow \infty} (\frac{2^{2k-3}+2^{k-2}}{3^{k-1}} - 1)^\frac{1}{k}[/itex]

After some calculations we got it reduced to

[itex] \lim_{k \rightarrow \infty} \frac{4}{3}(2^{\log_23-3}+2^{\log_23-2-k} - 2^{-k(2-\log_23)})^\frac{1}{k}[/itex]

Now, we belive this to be (4/3) but we get stuck with the limit

[itex]\lim_{k \rightarrow \infty} [\frac{3}{8} +\frac{3}{2^{k+2}} - \frac{3^k}{4^k}]^\frac{1}{k} [/itex]

Our thinking is that the last two terms within the brackets will be small so that this is somewhat like taking

[itex] \lim_{k\rightarrow \infty} (\frac{3}{8})^{\frac{1}{k}}=1 [/itex]?

and hence that the base tends to (4/3)? Empirically, this seems to be the case. However, how to you take this limit with mathematical rigour?

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# How do you take limit of k-roots when the terms within the root contains powers of k?

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