Problem source

Evaluate  the integral
\[
\hphantom{ \ \ \ \ \ \ \ \ \ 
(m> \tfrac12)\,.}
\int_{m-\frac12} ^\infty \frac 1{x^2}\,  \d x 
{ \ \ \ \ \ \ \ \ \ 
(m > \tfrac12)\,.}
\]
Show by means of a sketch that 
\[
\sum_{r=m}^n \frac 1 {r^2} 
\approx \int_{m-\frac12}^{n+\frac12} \frac1 {x^2} \, \d x
\,,
\tag{$*$}
\]
where $m$ and $n$ are positive integers with $m < n$.
\begin{questionparts}
\item You are given that the infinite series $\displaystyle \sum_{r=1}^\infty \frac 1 {r^2}$ converges to a value denoted by $E$. Use $(*)$ to obtain the following approximations for $E$:
\[
E\approx 2\,; \ \ \ \
E\approx \frac53\,; \ \ \ \
E\approx \frac{33}{20}
\,.\]
\item Show that, when $r$ is large,  the error in approximating $\dfrac 1{r^2}$  by $\displaystyle \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x$  is approximately $\dfrac 1{4r^4}\,$.
Given that $E \approx 1.645$, show that $\displaystyle \sum_{r=1}^\infty \frac1{r^4} \approx 1.08\, $.  
\end{questionparts}

Solution source

\begin{align*}
&& \int_{m-\frac12}^\infty \frac{1}{x^2} \d x &= \lim_{K \to \infty} \left [ -x^{-1} \right]_{m-\frac12}^K \\
&&&= \frac{1}{m-\frac12} - \lim_{K \to \infty }\frac{1}K \\
&&&= \frac{1}{m-\frac12}
\end{align*}


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/((#1)*(#1))};
    \def\xl{-1.5}; 
    \def\xu{11.5};
    \def\yl{-0.01}; \def\yu{.08};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) grid[step=1] (\xu,\yu);
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
    
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

         \foreach \x in {5,6,7,8,9,10} {
            \draw[curveB, fill=orange!40] ({\x-0.5}, 0) rectangle ({\x+0.5}, {\functionf(\x)});
        }
        
        \draw[curveA, domain=1:\xu, samples=150] 
            plot ({\x},{\functionf(\x)});

    \end{scope}
    
    % Annotate Function Names
    \node[curveA, labelbox] at ({(\xl+\xu)/2}, {\functionf((\xl+\xu)/2)}) {$y = x^{-2}$};
    % \node[curveB, labelbox] at (1.85, -1.1) {$x = \frac{4y^2+1}{3}$};
    

    \end{tikzpicture}
\end{center}

Notice that $\displaystyle \frac{1}{r^2} \approx \int_{r-\frac12}^{r+\frac12} \frac{1}{x^2} \d x$ as the area of the orange boxes and under the blue lines are similar.

\begin{questionparts}
\item $\,$
\begin{align*}
E &\approx \int_{1-\frac12}^\infty \frac1{x^2} \d x = \frac{1}{1-\frac12} = 2 \\
E &\approx 1 + \int_{2-\frac12}^\infty \frac1{x^2} \d x= 1 + \frac{1}{2 - \frac12} = \frac53 \\
E &\approx 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x= \frac54 + \frac{1}{3-\frac12} \\
&= \frac54+\frac{2}{5} = \frac{33}{20}
\end{align*}

\item The error is \begin{align*}
&& \epsilon &=  \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x - \frac1{r^2} \\
&&&= \frac{1}{r-\frac12} - \frac{1}{r + \frac12} - \frac1{r^2} \\
&&&= \frac{1}{r^2 - \frac14} - \frac1{r^2} \\
&&&= \frac{\frac14}{r^2(r^2-\frac14)} \\
&&&\approx \frac{1}{4r^4}
\end{align*}

Therefore \begin{align*}
&& \sum_{n=1}^\infty \frac1{r^4} &\approx 4 \left ( 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x-\sum_{r=1}^\infty \frac{1}{r^2}  \right)  + 1 + \frac{1}{2^4}\\
&&&= 4 \left ( \frac{33}{20}-1.645 \right) + 1 + \frac{1}{2^4}  \\
&&&=  4 \left ( 1.65-1.645 \right) + 1 + \frac{1}{2^4} \\
&&&= 1.0825 \approx 1.08
\end{align*}
\end{questionparts}