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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…

  • 0
    Well if 1 + 1/4 + ..... + 1/n^2 < 2 - 1/n^1 then 1+1/4 + ..... + 1/n^2 + 1/(n+1)^2 < 2 - 1/n^2 + 1/(n+1)^2 = 2 +(-(n+1)^2/n^2(n+1)^2 + n^2/(n+1)^2)= 2 +(-2n - 1)/(n+1)^2n^2 < 2 - 1/(n+1)^2$. So it follows by induction.2016-10-04

4 Answers 4

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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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    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.2011-05-05
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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: 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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Hint: Consider the bound $\dfrac{1}{n^2} < \dfrac{1}{n-1} - \dfrac{1}{n}$.

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    @user: I should also note that you should check [here](http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2) for methods of actually calculating the series.2011-05-05