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When practicing for my exam real analysis, I came across the following question:

"For each sequence $\{a_k\}$ of positive real numbers, we define $\{p_n\}$ by: $$ p_1=a_1, \qquad p_{n+1} = p_n a_n, \qquad \text{i.e. } \space p_n=a_1a_2...a_n. $$ Give an example of a sequence $\{a_k\}$ of positive real numbers such that: $$ \lim_{k\to \infty}a_k=1 \quad\text{ and} \quad \lim_{n\to \infty}p_n=0$$"

The solution, according to the book, is: $\space a_1=0 \space$ and $\space a_n=1 \space \space \forall n\geq1$.

However, I disagree with this since the question states that the sequence should consist of positive real numbers, not of non-negative real numbers (in the latter case the answer would have been fine of course). My own answer is

$$ a_n= {n\over n+1}$$

which is better in my opinion. But, I'd like to know whether my book's answer is really wrong, or if one could get away with including 0 in the positive reals?Thanks in advance!

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  • $\begingroup$ Yeah, $0$ is not a positive real number, so the solution given was wrong. Your solution is not just better, it is correct. :) $\endgroup$ – Thomas Andrews Apr 12 at 17:11
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    $\begingroup$ (I can only assume the answers in the book were written by someone else, because it also misses completely the point of the question, which is answers like yours.) $\endgroup$ – Thomas Andrews Apr 12 at 17:14
  • $\begingroup$ Some say positive but they mean non-negative.. They say strictly positive to mean positive. This world is confusing.. $\endgroup$ – Shashi Apr 12 at 17:16
  • $\begingroup$ Something is wrong with your recursion though, since the explicit form does not agree with the recursive form. Did you want to write $p_{n+1} = p_{n} a_{n+1}$? $\endgroup$ – Joker123 Apr 12 at 17:16
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    $\begingroup$ @Joker123 sorry, actually you are right. Did not see that until now. Then the question is also formulated wrong.. thanks for notifying me on that. $\endgroup$ – Benjamin Caris Apr 12 at 17:20

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