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Let $x_n$ be a sequence of the type described above. It is not monotonic in general, so boundedness won't help. So, it seems as if I should show it's Cauchy. A wrong way to do this would be as follows (I'm on a mobile device, so I can't type absolute values. Bear with me.) $$\left|x_{n+1} - x_n\right| \leq \frac{1}{n^2}.$$ So, we have $$ \left|x_m -x_n\right| \leq \sum_{k=n}^{m} \frac{1}{k^2}$$ which is itself Cauchy, etc., etc. But, of course, I can't just use absolute values like that. One thing I have shown is that $x_n$ is bounded. Inductively, one may show $$ \limsup_{n \to \infty} x_n \leq x_k + \sum_{k=n}^{\infty} \frac{1}{k^2},$$ although I'm not sure this helps or matters at all. Thanks in advance.

Disclaimer I've noticed that asking a large number of questions in quick succession on this site is often frowned upon, especially when little or no effort has been given by the asker. However, I am preparing for a large test in a few days and will be sifting through dozens of problems. Therefore, I may post a couple a day. I will only do so when I have made some initial, meaningful progress. Thanks.

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+1 for the disclaimer. – timur Aug 23 '12 at 19:35
Are you coming to penn? – neelp Aug 26 '12 at 1:51
I'm submatriculated. The questions you've asked here look very familiar... – user36387 Aug 26 '12 at 2:02
up vote 1 down vote accepted

Your last equation has a couple of bugs in it: you should write $\limsup$ instead of $\lim$ because you haven't yet shown that the limit exists, and the index of summation on the righthand side should match the index in the other RHS term.

Once you fix it up, though, it'll give you what you want. Since $\sum \frac{1}{k^2}$ converges, you can make the quantity $(\limsup_n x_n) - x_k$ as small as you like by choosing $k$ large enough.

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Oh, so I wasn't too far off. Thanks for the help. – user36387 Aug 23 '12 at 19:35

Note that $$ \lim_{n\to\infty}\sum_{k=n}^\infty\frac1{k^2}=0 $$ and for $m\gt n$, $$ a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ First, take the $\limsup\limits_{m\to\infty}$: $$ \limsup_{m\to\infty}a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ which must be non-negative. Then take the $\liminf\limits_{n\to\infty}$: $$ \limsup_{m\to\infty}a_m\le\liminf_{n\to\infty}a_n $$ Thus, the limit exists.

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I think the key is to use the liminf and limsup. Call them $a$ and $b$ respectively. Suppose they are not equal. Then their difference is some $3 \varepsilon > 0$ (You'll see why I multiplied by 3 in a moment). Now I need to use the fact that the series $\frac{1}{n^2}$ converges. Choose a natural number $i$ such that $\Sigma_{n=1}^{i} \frac{1}{n^2}$ is within $\varepsilon$ of it's limit ( I.E. $< \frac{6}{\pi ^2}$ for those who care).

From here we can say the following. Some subsequence of $x_n$ converges to $a$, so we can choose some natural $j$ such that $x_j$ is within $\varepsilon$ of the liminf. Now, the sequence of $x_n$ starting at $i+j$ is bounded above by $a+2\varepsilon$. In other words, always outside of $b$ by at least $\varepsilon$. This contradicts the fact that $b$ is the limsup. So $a=b$ thus the sequence converges.

Trying to do cauchy might be hard because it's difficult to exclude a large drop in the value of any particular $x_n$. Using the fact that the series converges is crucial. I'm pretty sure you can come up with a counter example using $\frac{1}{n}$ instead.

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