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I've tried to solve the below problem using the infinite geometric sequence formula but can't seem to equate it.

From given positive numbers, the following infinite sequence is defined: $a_1$ is the sum of all original numbers, $a_2$ is the sum of the squares of all original numbers, $a_3$ is the sum of the cubes of all original numbers, and so on ($a_k$ is the sum of the $k$-th powers of all original numbers).

Can it happen that $a_1 > a_2 > a_3 > a_4 > a_5$ while $a_5 < a_6 < a_7 <$ ...?

Can it happen that $a_1 < a_2 < a_3 < a_4 < a_5$ while $a_5 > a_6 > a_7 >$ ...?

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Choose the two positive numbers

$$a=0.8,\quad b=1.055.$$

The (unique) minimum of the function $a^n+b^n$ is at $n\approx 5.15885$. So yes, the first case can happen. The second case can not happen, because these functions are exponentials, hence convex and have no maximum.

This can also happen for arbitrary many numbers. Just choose all the other numbers to be $1$. They do not influence the extremums.

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  • $\begingroup$ What do you mean by convex? $\endgroup$ – juA May 11 '17 at 12:37
  • $\begingroup$ This means the function curves upwards ($f''>0$) like $e^x$ and not downwards ($f''<0$) like $\sqrt x$. See here. Exponentials are convex functions and such functions can only have minimums. $\endgroup$ – M. Winter May 11 '17 at 12:40
  • $\begingroup$ @juA Take $x_1,\dots,x_k$ to be the original numbers. The function $$ f(n) = x_1^n + x_2^n + \cdots + x_k^n $$ satisfies $a_n = f(n)$. Note that if the $x_k$ are positive, we must have $f''(n) > 0$, which means that $f$ can never have a maximum. $\endgroup$ – Omnomnomnom May 11 '17 at 12:45
  • $\begingroup$ @M.Winter whoops! Nice answer, then. $\endgroup$ – Omnomnomnom May 11 '17 at 12:49

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