December 16, 2024 · 9:19 am
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Interesting Sum

S=\sum_{n=1}^\infty (tan^{-1}(\frac{2}{n^2})), find S.

I came across this sum in An Imaginary Tale by Nahin and I was fascinated.

Let tan(\alpha)=n+1 and tan(\beta)=n-1.

Remember
tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}
Hence,
tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}
tan(\alpha-\beta)=\frac{2}{1+n^2-1}
tan(\alpha-\beta)=\frac{2}{n^2}
Therefore,
\alpha-\beta=tan^{-1}(\frac{2}{n^2})
and
\alpha=tan^{-1}(n+1) and \beta=tan^{-1}(n-1)

tan^{-1}(n+1)-tan^{-1}(n-1)=tan^{-1}(\frac{2}{n^2})

Which means,

    \begin{equation*}S=\sum_{n=1}^\infty(tan^{-1}(n+1)-tan^{-1}(n-1))\end{equation}

Let’s try a few partial sums

S_4=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)

S_4=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(4)+tan^{-1}(5)

S_6=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)+tan^{-1}(6)-tan^{-1}(4)+tan^{-1}(7)-tan^{-1}(5)

S_6=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(6)+tan^{-1}(7)

Hence, S_N=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(N)+tan^{-1}(N+1)

S_N=-\frac{\pi}{4}-0+tan^{-1}(N)+tan^{-1}(N+1)

What happens as N\rightarrow \infty ?

\lim\limits_{N\to \infty}\ S_N=-\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{3\pi}{4}

Because we know tan(\frac{\pi}{2}) is undefined.

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One Response to Interesting Sum

  1. Pingback: An Imaginary Tale – The story of i | Racquel Sanderson

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