Infinite series of (-1)^n+1sin(nx)/n and (-1)^n+1cos(nx)/n and Alternating Harmonic Series

        


            Today, I want to explore the series expansion of \log(1+x) in an analogous manner that we did in this post. Lets get started

\log(1+x)=\int_{0}^{x}\frac{du}{1+u}

=\int_{0}^{x}\sum_{n=0}^{\infty}(-1)^nu^ndu

=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{x}u^ndu

=\sum_{n=0}^{\infty}(-1)^n \, \Big[\frac{u^{n+1}}{n+1} \, \Big|_{0}^{x}\, \Big]

=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{n+1}

\boxed{\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n}(1)

Alternating Harmonic Series

The alternating harmonic series is given by

\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}

If we let x=1 in (1) we get

\boxed{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=\log 2}

Let´s now compute

S=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(2 \pi n x)


=\frac{1}{2 i}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \Big\{ e^{2 \pi i n x}-e^{-2 \pi i n x} \Big \}

=\frac{1}{2 i}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}e^{2 \pi i n x}}{n} -\frac{1}{2 i}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}e^{-2 \pi i n x} }{n}

=\frac{1}{2 i}\Big\{ \log(1+e^{2 \pi i x})-\log(1+e^{-2 \pi i x}) \Big \}

=\frac{1}{2 i}\Big\{ \log(1+e^{2 \pi i x})-\log(1+\frac{1}{e^{2 \pi i x}}) \Big \}

=\frac{1}{2 i}\Big\{ \log(1+e^{2 \pi i x})-\log(1+e^{2 \pi i x}) + 2 \pi i x \Big \}

=\frac{1}{2 i}\Big\{ \log \bigg(\frac{1+e^{2 \pi i x}}{1+e^{2 \pi i x}} \bigg) + 2 \pi i x \Big \}

=\frac{1}{2 i}\Big\{ \log(1) + 2 \pi i x \Big \}

=\frac{1}{2 i} 2 \pi i x

\boxed{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(2 \pi n x)=\pi x}

If we let  x=\frac{1}{2 \pi}  we get

{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin( n )}{n}=\frac{1}{2} }

Now we will compute

S=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\cos(2 \pi n x)

=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}e^{2 \pi i n x}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}e^{-2 \pi i n x}

=\frac{1}{2} \Big[ \log(1+e^{2 \pi i x})+ \log(1+e^{-2 \pi i x}) \Big]

=\frac{1}{2} \Big[ \log(1+e^{2 \pi i x})(1+e^{-2 \pi i x}) \Big]

=\frac{1}{2} \Big[ \log(2+e^{2 \pi i x}+e^{-2 \pi i x}) \Big]

=\frac{1}{2} \Big[ \log(2+2 \cos(2 \pi x) \Big]

Recalling the double angle formula for \cos(2A)

\boxed{\cos(2A)=1-2 \sin^2(A)}

we get

S=\frac{1}{2} \Big[ \log(2+2 (1-2\sin^2( \pi x)) \Big]

S=\frac{1}{2} \Big[ \log(4 -4\sin^2( \pi x)) \Big]

S=\frac{1}{2} \Big[ \log(4 (1-\sin^2( \pi x))) \Big]

S=\frac{1}{2} \Big[ \log(4 \cos^2( \pi x)) \Big]

S=\frac{1}{2} \Big[ \log(4)+ \log( \cos^2( \pi x)) \Big]


\boxed{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\cos(2 \pi n x)= \log(2)+ \log( \cos( \pi x)) }

Ricardo Albahari

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