The Fourier series expansion of the saw-toothed waveform f(x) = x in (- π, π) of period 2π gives the series, \(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots\)

The sum is equal to

Concept:

If f(x) is periodic function of period ‘’T’’ then f(x) can be expressed as below:

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \left( {nx} \right)\)

Where \({a_n} = \frac{2}{T}\mathop \smallint \nolimits_0^T f\left( x \right) \cdot \cos nx\;dx\)

\({b_n} = \frac{2}{T}\;\mathop \smallint \nolimits_0^T f\left( x \right) \cdot \sin nxdx\)

\({a_0} = \frac{1}{T}\;\mathop \smallint \nolimits_0^T f\left( x \right)dx\)

If f(x) is odd function, then only the coefficients of sin nx exists (i.e. a_n = 0 & a₀ = 0).

Calculation:

We are given f(x) = x for x ∈ (-π, π)

sin a f(x) = x is odd So, a_n = 0, a₀ = 0

So, now we have to find b_n,

\({b_n} = \frac{2}{{2\pi }}\mathop \smallint \nolimits_{ - \pi }^\pi x \cdot \sin \left( {nx} \right)dx = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi x \cdot {\rm{sin}}\left( {nx} \right)dx\)

\(\Rightarrow {{b}_{n}}=\frac{2}{\pi }\mathop{\int }_{0}^{\pi }x\cdot \sin \left( nx \right)dx~\{\because ~\mathop{\int }_{-a}^{a}f\left( x \right)=2\mathop{\int }_{0}^{a}f\left( x \right)~if~f\left( x \right)~is~odd\}\)

\(\Rightarrow {{b}_{n}}=\frac{2}{n}\cdot \left( -{{\left( -1 \right)}^{n}} \right)\)

\(f\left( x \right)=x=\frac{-2}{n}\cdot {{\left( -1 \right)}^{n}}\cdot \text{sin}\left( nx \right)\)

\(f\left( x \right)=x=2\sin x-\sin 2x+\frac{2}{3}\sin 3x-\frac{2}{4}\sin 2x+\ldots\)

Now put \(x=\frac{\pi }{2}\) to find \(f\left( \frac{\pi }{2} \right)\)

\(\Rightarrow f\left( \frac{\pi }{4} \right)=\frac{\pi }{2}=\left( 2 \right)~\left[ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots \right]\)

\(\Rightarrow 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}=\frac{\pi }{4}\)

Hence required sum of series is

Alternate Method:

Taylor expansion of tan^-1x is expressed in below:

\({{\tan }^{-1}}x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{{{x}^{7}}}{7}+\ldots for;~\left| x \right|<1\)

Put x = 9, above expansion series

\({{\tan }^{-1}}x={{\tan }^{-1}}1=\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\ldots\)