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Question Number 74109 by FCB last updated on 19/Nov/19

Commented by mr W last updated on 19/Nov/19

$f o r a_{k} > 0 a n d \underset{k = 1}{\sum^{n}} a_{k} = 1$ $w h e n a_{k} = \frac{1}{n},$ $m a x = {(n - 1)}^{n}$
Commented by MJS last updated on 20/Nov/19

$(\frac{1}{a_{n}} - 1) \underset{k = 1}{\prod^{n - 1}} (\frac{1}{a_{k}} - 1) \approx$ $[let a_{n} very close to 0 and a_{1} = a_{2} = . . . = a_{n - 1}]$ $\approx (\frac{1}{a_{n}} - 1) {(\frac{1}{\frac{1}{n - 1}} - 1)}^{n - 1} = (\frac{1}{a_{n}} - 1) {(n - 2)}^{n - 1}$ $n > 2 :$ $a_{n} \to 0^{+} \Rightarrow (\frac{1}{a_{n}} - 1) \to + \infty$ $the same Sir Aifour found before$
Commented by FCB last updated on 20/Nov/19

$thanks$
	Answered by ajfour last updated on 19/Nov/19

	$a_{1} \to 0, a_{2}, a_{3}, . . ., a_{n} \to \frac{1}{n - 1}$ $\Rightarrow (\to \infty) \times {(n - 2)}^{n - 1} \to \infty .$
	Commented by FCB last updated on 19/Nov/19

	$prove that sir$
	Commented by ajfour last updated on 19/Nov/19

	$n s h o u l d b e f i n i t e > 2 a n d a_{1} \to 0$ $c a n b e p o s s i b l e a s w e l l!$
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