How can we find sum of infinite terms of the series: $$\cot^{-1}(2)+\cot^{-1}\biggl(\frac{9}{2}\biggr)+\cot^{-1}(8)+\cot^{-1}\biggl(\frac{25}{2}\biggr)+\cot^{-1}(18)+ \dots$$

The difference of argument is in A.P. but it does not seem to be valuable. Could someone provide some hint?



The series can be written

$$\sum_{n=2}^\infty \text{arccot}\left(\frac{n^2}{2}\right) \tag 1$$

Then, note that

$$\text{arccot}(x)=\frac1x+O\left(\frac{1}{x^3}\right) \tag 2$$

as $x\to \infty$

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

From $(1)$ and $(2)$, we see that the series converges by comparison with $\sum_{n=2}^\infty \frac{1}{n^2}$. Then, we can write the summands as $$\text{arccot}\left(\frac{n^2}{2}\right)=\text{arccot}\left(\frac{n+2}{n}\right)-\text{arccot}\left(\frac{n}{n-2}\right)$$which provides a telescoping series. To evaluate the limit, we have $$\begin{align}\sum_{n=2}^N\left(\text{arccot}\left(\frac{n+2}{n}\right)-\text{arccot}\left(\frac{n}{n-2}\right)\right)&=\text{arccot}\left(\frac{N+2}{N}\right)\\\\&+\text{arccot}\left(\frac{N+1}{N-1}\right)\\\\&-\text{arccot}\left(3\right)\\\\&\to \frac{\pi}{2}-\text{arccot}(3)\,\,\text{as}\,\,N\to \infty\end{align}$$

  • $\begingroup$ This clearly gives the asymptotic behavior, but does it in fact yield the exact sum (if it converges)? $\endgroup$ – Brian Tung Apr 9 '16 at 19:14
  • $\begingroup$ @BrianTung It obviously converges by the comparison test. To find the sum, one needs to use the hint and telescope. $\endgroup$ – Mark Viola Apr 9 '16 at 19:23
  • $\begingroup$ Sorry, I knew it converged; I was trying not to divulge that. $\endgroup$ – Brian Tung Apr 9 '16 at 20:25
  • $\begingroup$ @BrianTung No need to apologize. The initial hint was not clear enough so I added a new section. -Mark $\endgroup$ – Mark Viola Apr 9 '16 at 20:47

Similarly as in Find sum of the Trignomertric series, you can use the identity $$ \DeclareMathOperator{\arccot}{arccot} \arccot(x) - \arccot(y) = \arccot\left(\frac{xy+1}{y-x}\right) $$ to conclude that $$ \arccot \frac{n+2}{n} - \arccot \frac{n}{n-2} = \arccot \frac{n^2}{2} $$ which allows to write your series as a sum of two telescoping series.


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.