# Show that $\lim\limits_{n\to\infty} x_n$ exists for $0 \le x_{n+1} \le x_n + \frac1{n^2}$

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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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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Why i can take $\varepsilon$ to zero? –  Knight Jun 30 '13 at 20:00
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 $L \leq \ell$. –  blabler Jun 30 '13 at 20:01
and why the sequence is bounded from below ? –  Knight Jun 30 '13 at 20:02
because $x_n \geq 0$ by assumptions –  blabler Jun 30 '13 at 20:03
my english is bad but the question is idk why exist L, i can not see the upper bound? –  Knight Jun 30 '13 at 20:05

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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This only shows that if the limit exists, then it is bounded between zero and $x_1 + \pi^2/6$ –  blabler Jul 2 '13 at 17:00
Yes, thanks @blabler, I should edit my answer. –  Samrat Mukhopadhyay Jul 3 '13 at 9:33