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I am solving this problem. If $\sum\limits_{i=1}^{\infty} \tan^{-1}\biggl(\frac{1}{2i^{2}}\biggr)= t$ then find the value of $\tan{t}$.

My solution is like the following: I can rewrite: \begin{align*} \tan^{-1}\biggl(\frac{1}{2i^{2}}\biggr) & = \tan^{-1}\biggl[\frac{(2i+1) - (2i-1)}{1+(2i+1)\cdot (2i-1)}\biggr] \\\ &= \tan^{-1}(2i+1) - \tan^{-1}(2i-1) \end{align*}

and when I take the summation the only term which remains is $-\tan^{-1}(1)$, from which I get the value of $\tan{t}$ as $-1$. But the answer appears to be $1$. Can anyone help me on this.

4 Answers 4

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By telescopy we deduce

$\begin{array}{c l}\sum_{k=1}^\infty \tan^{-1}\left(\frac{1}{2k^2}\right) & = \lim_{n\to\infty}\sum_{k=1}^n\tan^{-1}\left(\frac{1}{2k^2}\right) \\ & = \lim_{n\to\infty}\sum_{k=1}^n\left[\tan^{-1}(2k+1)-\tan^{-1}(2k-1)\right] \\ & =\lim_{n\to\infty}\left[\color{Purple}{\tan^{-1}(2n+1)}-\tan^{-1}(1)\right]. \end{array}$

However, the term in purple does not tend to zero as $n\to\infty$, it tends to something else...

  • 0
    FWIW, I can barely see the distinction between black and purple on my screen, and it is generally unwise to indicate information with color alone because enough of the human population is colorblind. (Not me, I'm just old.) It's okay to duplicate information with color (making it clear with text what you are talking about, but making it visually striking to the readers who can see the color) but it should rarely be the only way you present information.2017-05-09
7

How about trying the same identity, but using the fact that $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan(x)}$. $ \begin{align} \sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k^2}\right) &=\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k-1}\right)-\tan^{-1}\left(\frac{1}{2k+1}\right)\\ &=\tan^{-1}(1)\\ &=\frac{\pi}{4} \end{align} $ Of course, then $\tan\left(\frac{\pi}{4}\right)=1$

5

In this answer, it is shown that using the equation $ \frac1{2k^2}=\frac{\frac1{2k-1}-\frac1{2k+1}}{1+\frac1{2k-1}\frac1{2k+1}} $ and the identity $ \tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)} $ we get $ \begin{align} \sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k^2}\right) &=\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k-1}\right)-\tan^{-1}\left(\frac{1}{2k+1}\right)\\ &=\tan^{-1}(1)\\ &=\frac{\pi}{4} \end{align} $

  • 0
    Really, you should be stopping at the step $\tan^{-1}1$, since the question was to find $\tan t$, which is $1$. (i.e. the calculation $\tan^{-1}1=\frac\pi4$ is irrelevant)2014-04-20
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HINT:

$\frac1{2n^2}=\frac2{1+4n^2-1}=\frac{2n+1-(2n-1)}{1+(2n+1)(2n-1)}$

$\arctan x-\arctan y=\arctan\left(\frac{x-y}{1+xy}\right)$

  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/14018/discussion-between-user34304-and-lab-bhattacharjee)2014-04-20