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One can prove the following statement:

Let
$(1) a_1, a_2, a_3, ... a_n, ...$
be a sequence of non-negative numbers. Let also $k>1$ be an integer.

If the sequence
$(2) a_1^k, a_2^k, a_3^k, ... a_n^k, ...$
converges and its limit is $a$, then the sequence (1) also converges and its limit is $\sqrt[k]{a}$.

But what if instead of (2) we know that the sequence
$(3) a_1^1, a_2^2, a_3^3, ... a_n^n, ...$
converges and its limit is b. Can we then state something about (1), and about its convergence, and possibly about its limit (if such a limit exists)?

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1 Answer 1

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Take $(a_n)$ defined by :

$a_n = 0$ if n is even.

$a_n = \frac{1}{2}$ if n is odd.

This sequence satisfies (3) but it does not converge.

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  • $\begingroup$ Hm, in that case, can we add something, some additional condition to my statement to make it valid? I guess not (or not so easy) but ... Your example just make me think further. In any case, thanks for this example. $\endgroup$ Commented Mar 27, 2015 at 10:54
  • $\begingroup$ @peter.petrov as I pointed out in the comments in Surb answer, the convergence of $(a_n^n)_{n\in \mathbb{N}}$ with the given setting only grants you that every accumulation Point is strict smaller 1 $\endgroup$ Commented Mar 27, 2015 at 10:57
  • $\begingroup$ @DominicMichaelis OK, thanks, I will think some more about it. $\endgroup$ Commented Mar 27, 2015 at 10:58
  • $\begingroup$ @peter.petrov sorry got a flaw, even $1$ may be an accumulation Point of the sequence. $\endgroup$ Commented Mar 27, 2015 at 11:01

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