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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

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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

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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.

Here is an approach

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

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