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I have doubt in this answer,

Unfortunately, the choice $e^{1/e}$ was not given. The two values of $x$ that you are given that bracket $e$ are $2$ and $3$, so the correct choice is either $2^{1/2}$ or $3^{1/3}$.

How does one prove that if we extrapolate a sequence $a_n$ as a function $f(x)$ and extrema of $f(x)$ occurs at $x=j$ ( $j \in [0, \infty]$) then the extrema of the sequence occurs at the whole numbers which bracket $j$?

tryst with freedom
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  • "hen the extrema of the sequence occurs at the whole numbers which bracket $j$"? Why do you think this is true? In THAT answer, $2^{1/2}$ and $3^{1/3}$ are the closest options to $e^{1/e}$. Thats why you only need to take them. Well, BEFORE e, the maximum vlaue is $2^{1/2}$ and AFTER $e$, $3^{1/3}$. – Tito Eliatron Jun 14 '21 at 16:13
  • Right, but the $ e \in \left[ 2,3 \right]$, so it means the index which corresponds to bracket of $e$ is 2 and 3. Hence the idea in the statement? – tryst with freedom Jun 14 '21 at 16:14
  • @solublefish, I meant that for $a_n = n^{\frac{1}{n} } $ then our function is $f(x) = x^{\frac{1}{x} }$ , it must be that function is similar to sequence – tryst with freedom Jun 14 '21 at 17:22

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