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$$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$$

We know that : $\arctan{x} - \arctan{y} = \arctan{\frac{x-y}{1+xy}}$ for every $ xy > 1 $

I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.

Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$

Similarly here :

$$\sum\limits_{n=1}^{\infty}\arctan{\frac{8n}{n^4-2n^2+5}}$$

The result should be $ \arctan 2 $ on the first one and $ \pi/2 + \arctan2 $ on the second one.

lab bhattacharjee
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2 Answers2

9

Hint:

$$\dfrac2{n^2+n+4}=\dfrac{\dfrac12}{1+\dfrac{n(n+1)}4}=\dfrac{\dfrac{n+1}2-\dfrac n2}{1+\dfrac n2\cdot\dfrac{n+1}2}$$

For the second, $$\dfrac{8n}{n^4-2n^2+5}=\dfrac{\dfrac{(n+1)^2}2-\dfrac{(n-1)^2}2}{\dfrac{(n+1)^2}2\cdot\dfrac{(n-1)^2}2+1}$$

lab bhattacharjee
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    See also: https://math.stackexchange.com/questions/193001/explicitly-finding-the-sum-of-arctan1-n2n1 https://math.stackexchange.com/questions/2789725/evaluating-tan-left-sum-r-1-infty-arctan-left-frac44r2-3-right – lab bhattacharjee Jan 28 '19 at 11:58
  • Can you explain in more details how the expressions for $a$ and $b$ in both cases were found? – user Feb 19 '19 at 12:27
  • @user, Where are $a,b$ in the current version ? – lab bhattacharjee Feb 19 '19 at 12:53
  • @user, https://math.stackexchange.com/questions/3107476/if-sum-n-1-infty-tan-1-left-frac4n2n16-right-tan-1-left-frac/3109078#3109078 – lab bhattacharjee Feb 19 '19 at 13:09
  • Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=\arctan(an^2+bn+c)$ as a try function? – user Feb 19 '19 at 13:20
  • @user, Observe that in the second case we have $$f(n+1)-f(n-1)$$ But in https://math.stackexchange.com/questions/3107476/if-sum-n-1-infty-tan-1-left-frac4n2n16-right-tan-1-left-frac/3109078#3109078,we have assumed that $\arctan$ is expressible in the form $$f(n+1)-f(n)$$ – lab bhattacharjee Feb 19 '19 at 14:53
  • I have noted this. Therefore I was wondering if there is a general method to find $f(n)$ and $k$ satisfying the equation $\frac{f(n+k)-f(n)}{1+f(n+k)f(n)}=F(n)$ where $F(n)$ is a rational functon of $n$. – user Feb 19 '19 at 16:09
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$$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}} = \sum\limits_{n=1}^{\infty}\arctan{\frac{2(n+1)-2n}{n(n+1)(1+\frac{4}{n(n+1)})}} = \sum\limits_{n=1}^{\infty}\arctan{\frac{\frac{2}{n}-\frac{2}{n+1}}{1+\frac{2} {n}\cdot\frac{2}{n+1}}} =\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n}-\arctan{\frac{2}{n+1}}} =\arctan2$$

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