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I encountered the series $$ \sum_{n=1}^{\infty} \arctan\frac{2}{n^{2}}. $$

I know it converges (by ratio test), but if I need to calculate its limit explicitly, how do I do that? Any hint would be helpful..

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    Wolfy says 3pi/4 – 2017-02-20
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    Dupl http://math.stackexchange.com/questions/243802/showing-that-sum-n-1-infty-arctan-left-frac2n2-right-frac3 – 2017-02-20

2 Answers 2

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Note that $\arctan(u)-\arctan(v)=\arctan\left(\frac {u-v}{1+uv}\right)$. Taking $u=n+1$ and $v=n-1$ shows that $$\arctan\left(\frac {2}{n^2}\right)=\arctan(n+1)-\arctan(n-1)$$

Thus we see that the series telescopes and $$\sum_{n=1}^{\infty}\arctan\left(\frac {2}{n^2}\right)=2\arctan(\infty)-\arctan(0)-\arctan(1)=\pi -0-\frac {\pi}4=\frac {3\pi}4$$

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\begin{align*} \sum_{n=1}^\infty\arctan\left ( \frac{2}{n^2} \right ) &=-arg \prod_{n=1}^\infty\left (1-\frac{2i}{n^2} \right ) \\ &=-arg \prod_{n=1}^\infty\left (1-\frac{(\sqrt{2i})^2}{n^2} \right ) \\ &=-arg\left(\frac{\sin(\pi\sqrt{2i})}{\pi\sqrt{2i}} \right ) \\ &=-arg\left(-\frac{(1/2+i/2)\sinh\left(\pi \right )}{\pi} \right ) \\ &= \frac{3\pi}{4} \end{align*}

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    @ep pi please explain how you get $\displaystyle \arg \prod^{\infty}_{n=1}\bigg(1-\frac{2i}{n^2}\bigg)$ and also explain second last line., Thanks – 2017-02-27
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    @Durgesh Tiwari The inspiration for this infinite product comes from $arg\;\left(1-\frac{2i}{n^2}\right)=-\arctan(2/n^2)$ and $arg\prod_n r_n e^{\theta_n}=\sum_n\theta_n$. What I did in the second to last line occurs due to the branch I had chosen to work on. – 2017-03-03
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    I have made an error in my previous comment, it should be $r_n e^{i\theta_n}%$ under the product, not $r_n e^{\theta_n}$. – 2017-03-03