7
$\begingroup$

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..

| cite | improve this question | | | | |
$\endgroup$
10
$\begingroup$

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 $$\sum_{n=1}^{\infty}\arctan\left(\frac {2}{n^2}\right)=2\arctan(\infty)-\arctan(0)-\arctan(1)=\pi -0-\frac {\pi}4=\frac {3\pi}4$$

| cite | improve this answer | | | | |
$\endgroup$
3
$\begingroup$

\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*}

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ @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 $\endgroup$ – DXT Feb 27 '17 at 9:56
  • 1
    $\begingroup$ @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. $\endgroup$ – dxdydz Mar 3 '17 at 0:20
  • 1
    $\begingroup$ I have made an error in my previous comment, it should be $r_n e^{i\theta_n}%$ under the product, not $r_n e^{\theta_n}$. $\endgroup$ – dxdydz Mar 3 '17 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.