Prove that $\exists c>0$ s.t $\sum_{n\geq x}\frac{1}{n^2}\leq \frac{c}{x}$

Question

$\mathbf{Question:}$ Prove that there exists a constant $c>0$ such that for all $x \in [1, \infty)$, $\displaystyle\sum_{n \geq x}\frac{1}{n^2} \leq \frac{c}{x}$

$\mathbf{Attempt:}$ Evidently,

$\displaystyle\bigg[\displaystyle\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]{\lfloor x \rfloor}\leq x\displaystyle\sum_{n \geq x}\frac{1}{n^2} \leq \displaystyle\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]{\lceil x \rceil} $.

This inequality arises since,

$\displaystyle\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]=\displaystyle\sum_{n \geq x}\frac{1}{n^2}$.

Now, let $\lfloor x\rfloor=m$

$\displaystyle\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg]m$

$=\lim_{{m \to \infty},{r\to \infty}}\bigg[\frac{m}{(m+1)^2}+\frac{m}{(m+2)^2}+\frac{m}{(m+3)^2}+...+\frac{m}{(m+r)^2}+...\bigg]=$

$\lim_{{h \to 0},{r\to \infty}}h\bigg[\frac{1}{(1+h)^2}+\frac{1}{(1+2h)^2}+\frac{1}{(1+3h)^2}+...+\frac{1}{(1+rh)^2}+...\bigg]=\displaystyle\int_{x=0}^\infty\frac{1}{(1+x)^2}dx=1$

We have, $\lceil x\rceil=m+1$ and $\displaystyle\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg](m+1)=\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg]m$

Thereby, we conclude, $\displaystyle \lim_{x\to \infty}x\sum_{n \geq x}\frac{1}{n^2}=1 $.

Therefore, from the definition of limit, $\forall \varepsilon>0$, $\exists G>0$ such that $\bigg|x\displaystyle\sum_{n \geq x}\frac{1}{n^2}-1\bigg|<\varepsilon$ $\forall x>G$. Choosing $\varepsilon =1$, we get $G=G_0$ such that $x\displaystyle\sum_{n \geq x}\frac{1}{n^2}<2$, for any $x> G_0$.

We set $\displaystyle c=\max\bigg\{2, \sup_{x\in [1,G_0]}x\sum_{n \geq x}\frac{1}{n^2} \bigg\}$

Is this procedure valid? Kindly verify.

Answer 1

Comments

I had initially thought that, although $x$ is usually a real variable, $x$ was intended to be an integer. However, if $x$ is intended to take on any real value, one needs to be clear on exactly what is meant, since this is not a standard use for $\sum$. Here are two that come to mind at first: $$ \sum_{n\ge x}\frac1{n^2}=\sum_{n=\lceil x\rceil}^\infty\frac1{n^2} $$ and $$ \sum_{n\ge x}\frac1{n^2}=\sum_{n=0}^\infty\frac1{(n+x)^2} $$ That point aside, your approach of finding the limit and using that for all $x$ past a finite value, and then handling the initial finite range of $x$ separately, is quite valid.

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Slick Answer

For $m\ge1$, $$ \begin{align} \sum_{n\ge m}\frac1{n^2} &\le\sum_{n\ge m}\frac1{\left(n-\frac12\right)\left(n+\frac12\right)}\\ &=\sum_{n\ge m}\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right)\\ &=\frac1{m-\frac12}\\[3pt] &\le\frac2m \end{align} $$

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Less Slick Answer

For $m\ge2$, $$ \begin{align} \sum_{n\ge m}\frac1{n^2} &\le\sum_{n\ge m}\frac1{(n-1)n}\\ &=\sum_{n\ge m}\left(\frac1{n-1}-\frac1n\right)\\ &=\frac1{m-1}\\[3pt] &\le\frac2m \end{align} $$ For $m=1$, $$ \begin{align} \sum_{n\ge m}\frac1{n^2} &\le1+\sum_{n\ge 2}\frac1{(n-1)n}\\ &=1+\sum_{n\ge 2}\left(\frac1{n-1}-\frac1n\right)\\ &=2\\[6pt] &=\frac2m \end{align} $$

Answer 2

I think you would be better off starting with an integral from the start: $$\frac 1{n^2} \le \int_{n-1}^{n} \frac 1{t^2} \, dt.$$ For any integer $k$ [edited to address complaint] integer $k \ge 2$ you have $$\sum_{n=k}^\infty \frac 1{n^2} \le \int_{k-1}^\infty \frac 1{t^2} \, dt = \frac 1{k-1}$$

From here it is not too difficult.