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I havent done this in a while so I was hoping someone can remind me how to do this,

I need to find the limit of this summation:

$$\lim_{n \to \infty}{\displaystyle\sum_{k=1}^{n} \frac{1}{k^2}} $$

How exactly would I do this?

The reason I'm asking this is because im actually trying to prove $${{\displaystyle\sum_{k=1}^{n} \frac{1}{k^2}} \in \theta(1)}$$

so i was thinking I could use limits to prove $${\exists c, c' \in R^+}, {n{\scriptstyle 0} \in N}: cf(n){\le} g(n) {\le} c'f(n)$$

where g(n) is the summation and f(n) is 1

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marked as duplicate by ncmathsadist, Zev Chonoles Feb 10 '13 at 21:27

This question was marked as an exact duplicate of an existing question.

Duplicate of – Zev Chonoles Feb 10 '13 at 21:23
If you merely want to show that the series is bounded, use $\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ and notice that the latter telescopes. – Ayman Hourieh Feb 10 '13 at 22:24

Do you mean the limit as $n \to \infty$? The result, $\frac{\pi^2}{6}$, is well known but AFAIK there isn't an elementary proof. You'll need something a little more advanced like Fourier series to do it. Depending on what you need this for this might be a summation that your professor just expects you know.

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there's also a proof using complex analysis methods. – user59671 Feb 10 '13 at 21:34

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