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I'm currently taking a Comp Sci class that is reviewing Calculus 2. I have a question:

Show that the summation $\sum_{i=1}^{n}\frac{1}{i^2}$ is bounded above by a constant

I realize that this question is already answered here

Showing that the sum $\sum_{k=1}^n \frac1{k^2}$ is bounded by a constant

Could anyone explain it to me further? I believe I'm supposed to use p-series test or integral test to complete

Community
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user1781500
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    I think the answer in that link is pretty clear... – DonAntonio Jan 29 '14 at 04:33
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    Can you clarify what you don't understand about the linked answer? –  Jan 29 '14 at 04:34
  • @anorton, I highly doubt this question would be interesting in the least for anyone that could "say" that the infinite series gotten from the finite one by taking the limit $;n\to\infty;$ is $;\zeta(2);$ ... – DonAntonio Jan 29 '14 at 04:42
  • You seem to have knowledge of the integral test - why didn't you simply apply that? It shows very quickly that the sum converges, and hence is bounded above. – G. H. Faust Jan 29 '14 at 05:07
  • If you are interested also in the value of the sum, see here: http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-n-1-infty-frac1n2 – Martin Sleziak Jan 29 '14 at 08:46
  • These question are also related: http://math.stackexchange.com/questions/58931/does-sum-limits-k-1n-1-k-2-converge-when-n-rightarrow-infty and http://math.stackexchange.com/questions/333417/need-to-prove-the-sequence-a-n-1-frac122-frac132-cdots-frac1n – Martin Sleziak Jan 29 '14 at 08:48
  • @anorton: No biggie! – Ted Shifrin Jan 29 '14 at 19:21

2 Answers2

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You can try the following, pretty similar to that link's, approach:

$$\frac1{k^2}\le\frac1{k(k-1)}=\frac1{k-1}-\frac1k\implies$$

$$\sum_{k=2}^N\frac1{k^2}\le\sum_{k=2}^N\left(\frac1{k-1}-\frac1k\right)=1-\frac12+\frac12-\frac13+\frac13-\frac14+\ldots+\frac1{N-1}-\frac1N=$$

$$=1-\frac1N\le 1$$

and you've proved boundedness for any $\;N\in\Bbb N$

Notes: Observe that in the above we take the sum from $\;n=2\;$ and not, as you did, from $\;n=1\;$ . Answer:

(1) Why?

(2) Explain why the above doesn't affect the boundedness of the original series.

DonAntonio
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Here is an approach

$$ \sum_{k=1}^{n}\frac{1}{i^2} = 1 + \sum_{k=2}^{n}\frac{1}{i^2} \leq 1 + \int_{1}^{n}\frac{dx}{x^2}= 2 - \frac{1}{n}\leq 2,\, \quad \forall n\in \mathbb{N}.$$

Mhenni Benghorbal
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