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I need to show that, for every $n\in\mathbb{N}$, we obtain:

$$ 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq2. $$

My proof is the following: We know that $\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}$, so

$$ \sum_{i=1}^n\frac{1}{i^2}\leq\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}<2. $$

But I want to know if exists another proof for this problem. I try induction, but I can't do it.

Thanks for advance.

Michael Rozenberg
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YCB
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4 Answers4

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$$\sum_{k=1}^n\frac{1}{k^2}\leq1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=2-\frac{1}{n}<2$$

Michael Rozenberg
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  • This is clearly a nice way of doing it. If one would like not to use this trick with the telescoping argument, one could try to show that the sum is bounded by $2-1/n$ instead of $2$. That will simply work out better. So, as it turns out, it is sometimes easier to show a stronger result by induction. – mickep Jan 25 '18 at 19:04
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\begin{align*} \sum_{k=1}^{n}\dfrac{1}{k^{2}}=1+\sum_{k=2}^{n}\dfrac{1}{k^{2}}<1+\int_{1}^{\infty}\dfrac{1}{t^{2}}dt=2. \end{align*}

user284331
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Hint: Telescopic sum and $$\frac{1}{k^2} \le \frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.$$

Math Lover
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Let's get creative. We have $$\zeta(2)= \sum_{n\geq 1}\frac{1}{n^2} =\sum_{n\geq 1}\iint_{(0,1)^2}x^{n-1}y^{n-1}\,dx\,dy=\iint_{(0,1)^2}\frac{dx\,dy}{1-xy}$$ and the integral $$ \iint_{(0,1)^2}\frac{x^2(1-x)^2 y^3(1-y)^3}{1-xy}\,dy\,dx \stackrel{}{=}\frac{329}{40}-5\,\zeta(2)$$ is small but clearly positive, hence $\zeta(2)\leq\color{red}{\frac{329}{200}}=1.645$. Similarly, $$ \iint_{(0,1)^2}\frac{x^2(1-x)^2 y^2(1-y)^2(x-y)^2(x+y-1)^2}{1-xy}\,dx\,dy =\frac{25003}{400}-38\,\zeta(2) $$ leads to $\zeta(2)\leq \color{red}{\frac{25003}{15200}}=1.64493421\ldots$

Jack D'Aurizio
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