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Let $x_1, x_2,\ldots$ be a sequence of non-negative real numbers such that

$$ x_{n+1} ≤ x_n + \frac 1{n^2}\text{ for }1≤n. $$

Show that $\lim\limits_{n\to\infty} x_n$ exists. Help please...

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2 Answers 2

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The sequence is bounded from above and below, hence both $$ \ell = \liminf x_n $$ and $$ L = \limsup x_n $$ are finite. Pick $\varepsilon$ and a very large $n$ at which we have both $$x_n < \ell + \varepsilon$$ and $$\sum_{k \geq n} \frac{1}{k^2} < \varepsilon$$ Then for any $m > n$ using the assumption, we get $$ x_m < x_n + \sum_{k \geq n} \frac{1}{k^2} \leq \ell + 2\varepsilon $$ Taking $m$ to infinity along a sequence such that $x_m \rightarrow L$ we get $L \leq \ell + 2\varepsilon$. Taking $\varepsilon$ to zero we get $L \leq \ell$. Since trivially $\ell \leq L$ we conclude that $\ell = L$ and therefore the limit exists.

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  • $\begingroup$ Why i can take $\varepsilon$ to zero? $\endgroup$ Jun 30, 2013 at 20:00
  • $\begingroup$ You picked $\varepsilon > 0$ arbitrary, and then proved that $L \leq \ell + 2\varepsilon$. This is an inequality between $L$ and $\ell$ that holds for every $\varepsilon > 0$ so you might as-well take $\varepsilon \rightarrow 0$ and then get $L \leq \ell$. $\endgroup$
    – blabler
    Jun 30, 2013 at 20:01
  • $\begingroup$ and why the sequence is bounded from below ? $\endgroup$ Jun 30, 2013 at 20:02
  • $\begingroup$ because $x_n \geq 0$ by assumptions $\endgroup$
    – blabler
    Jun 30, 2013 at 20:03
  • $\begingroup$ my english is bad but the question is idk why exist L, i can not see the upper bound? $\endgroup$ Jun 30, 2013 at 20:05
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From the given condition and equation, it follows$$0\leq x_{n+1}\leq x_1 +\sum_{k=1}^{n}\frac{1}{n^2}$$ Taking limits as $n\rightarrow \infty$ we get $$0\leq \lim_{n\rightarrow \infty}x_{n}\leq x_1+\sum_{n=1}^{\infty}\frac{1}{n^2}=x_1+\zeta(2)=x_1+\frac{\pi^2}{6}$$

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    $\begingroup$ This only shows that if the limit exists, then it is bounded between zero and $x_1 + \pi^2/6$ $\endgroup$
    – blabler
    Jul 2, 2013 at 17:00
  • $\begingroup$ Yes, thanks @blabler, I should edit my answer. $\endgroup$ Jul 3, 2013 at 9:33

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