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Prove that for any nonzero natural $n$ it is true that $$S_n = 1 + 1/4 + 1/9 + 1/16 + 1/25 + … + 1/n^2 < 2.$$

I'm sort of at a loss here. I'm not sure if there exists some formula or method to sum this kind of series, since there is a variable ratio…

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Please do not post in the imperative. If you have a question, please ask. Also, what have you tried? –  JavaMan May 5 '11 at 4:34
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You don't seem to have a series... –  Mariano Suárez-Alvarez May 5 '11 at 4:35
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Moreover, your statement is clearly false: for example, if $n=1$ then your inequality does not hold. –  Mariano Suárez-Alvarez May 5 '11 at 4:36
    
People are answering a question that is not the one asked... (or, the challenge posed, rather) –  Mariano Suárez-Alvarez May 5 '11 at 4:39
    
@Mariano. You are right. I have deleted my posts. –  JavaMan May 5 '11 at 4:40
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4 Answers

For $n>1$, a formula that will help is $$\frac{1}{n^2}<\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}.$$ This gives a telescoping series as an upper bound.

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So I should just take the limit of (1/(n-1) - 1/n) as n approaches infinity? –  user10504 May 5 '11 at 4:58
    
No, you should see what happens if you use this upper bound. For example, $1+\frac{1}{4}<1+1-\frac{1}{2}$. $1+\frac{1}{4}+\frac{1}{9}<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}$. Simplify, continue the pattern and see what develops. You may also want to search for references to telescoping series. –  Jonas Meyer May 5 '11 at 5:03
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Hint: If you replace the $\frac{1}{n^2}$ terms after the first with the greater $\frac{1}{n(n-1)}$, you can use partial fractions and telescope the series. Alternately, there are difficult proofs that your series sums to $\frac{\pi^2}{6}\approx 1.64493 \lt 2$

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you can compare with $1+\int_1^{\infty}n^{-2}$ (draw a picture) which is exactly $2$. then estimate any little bit of the error to get below $2$

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Hint: Consider the bound $\dfrac{1}{n^2} < \dfrac{1}{n-1} - \dfrac{1}{n}$.

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Unfortunately, 2^n>(n+1)^2 for n large enough, so this does not bound the series. –  Ross Millikan May 5 '11 at 4:40
    
Oh my gosh - you're right! Instead, consider the classic telescopic series! I'll edit my post. –  mixedmath May 5 '11 at 4:42
    
@user: I should also note that you should check here for methods of actually calculating the series. –  mixedmath May 5 '11 at 4:44
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