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Suppose that $\{x_n\}_{n=1}^{\infty}$ is a bounded sequence, and that $x_n>0$ holds for all positive integer $n$.

Find $\lim_{n \to \infty} \frac{x_n}{x_1 + \cdots + x_n}$.

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    $\begingroup$ Any ideas, guesses, thoughts, work? $\endgroup$ – blue Sep 2 '14 at 3:23
  • $\begingroup$ Is it series or sequence you meant? $\endgroup$ – Anjan3 Sep 2 '14 at 3:52
  • $\begingroup$ @AnjanDebnath does it affect your understanding? $\endgroup$ – John Fernley Sep 2 '14 at 4:17
  • $\begingroup$ @JohnFernley Does my understanding affect your understanding? For the first time, when I saw the question it was " Suppose that $\{x_n\}_{n=1}^\infty$ is a bounded 'series' and now after the editing is done(after you made the comment), I see "bounded sequence". Do I need to elaborate more for understanding? $\endgroup$ – Anjan3 Sep 2 '14 at 5:10
  • $\begingroup$ @AnjanDebnath If you understand the question don't distract from it with this pedantry. We weren't talking about my understanding. $\endgroup$ – John Fernley Sep 2 '14 at 21:07
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Hint: It is useful to separately consider the cases $\sum x_n < \infty$ and $\sum x_n = \infty$. In the latter case, remember that $\{x_n\}$ is bounded.

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  • $\begingroup$ Do you know that the limit exists ? $\endgroup$ – Rene Schipperus Sep 2 '14 at 3:32
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    $\begingroup$ @Rene Do you mean the sum? The partial sums are monotonic increasing, so either they are unbounded and diverge to $+\infty$ or are bounded and hence have a limit. $\endgroup$ – blue Sep 2 '14 at 3:38
  • $\begingroup$ Sorry, you are right about that, but if the series diverges, I dont see that the limit exists or is even uniquely determined. $\endgroup$ – Rene Schipperus Sep 2 '14 at 3:40
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    $\begingroup$ @Rene When a numerator is bounded and the denominator monotonically increases without bound, of course the ratio's limit exists! $\endgroup$ – blue Sep 2 '14 at 3:45
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I think the limit is zero. The above hint is useful. For $\sum x_n<\infty$, $x_n\rightarrow 0$ ($n\rightarrow \infty$). In the case $\sum x_n=\infty$, the answer is obvious.

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    $\begingroup$ Boundedness of the sequence is required in the second part, I won't hear "obvious". $\endgroup$ – John Fernley Sep 2 '14 at 4:16

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