Problem source

Give a rough sketch of the function $\tan^{k}\theta$ for $0\leqslant\theta\leqslant\frac{1}{4}\pi$
	in the two cases $k=1$ and $k\gg1$ (i.e. $k$ is much greater than
	1). 

	Show that for any positive integer $n$ 
	\[
	\int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right),
	\]
	and deduce that 
	\[
	\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2.
	\]
	Show similarly that 
	\[
	\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}.
	\]

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-0.5};
    \def\xu{1.2};
    \def\yl{-0.5};
    \def\yu{1.2};
    
    % 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 styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:{pi/4}, samples=100] 
            plot (\x, {tan(deg(\x))});
        \node[blue] at (0.5,{tan(deg(0.5))}) {$y=\tan x$};
        \draw[thick, red, smooth, domain=0:{pi/4}, samples=100] 
            plot (\x, {tan(deg(\x))^5});
        \node[red] at (0.62,{tan(deg(0.62))^5}) {$y=\tan^k x$};
        
    \end{scope}


    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

Let $\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta$, then

\begin{align*}
I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\
&= \left [ -\ln \cos \theta  \right]_0^{\pi/4} \\
&= -\ln \frac{1}{\sqrt{2}} - 0 \\
&= \frac12 \ln 2 \\
\\
\\
I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\
&= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\
&= \int_0^{\pi/4}  \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\
&= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\
&= \frac1{2n} - I_{2n-1}
\end{align*}

Therefore we can see that $\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)$.

As we can see as $n \to \infty$, $I_n \to 0$

Therefore \begin{align*}
&& 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\
\Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}
\end{align*}

\begin{align*}
&& I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\
&&&= \frac{\pi}{4}
\end{align*}

Therefore $\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)$ and since $I_{2m} \to 0$ the same result follows.