Question Number 227051 by mnjuly1970 last updated on 28/Dec/25
$$ \\ $$$$\:\:\:\:\:\underset{{n}\in\mathbb{N}\cup\left\{\mathrm{0}\right\}} {\sum}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}\:+\:\mathrm{2}}\:\right)=?\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Spillover last updated on 28/Dec/25
$$\frac{\pi}{\mathrm{4}}? \\ $$
Answered by Spillover last updated on 28/Dec/25
![Rewritten using partial fractions (1/(n^2 +3n+2))=(1/(n+1))−(1/(n+2)) tan^(−1) ((1/(n+1))−(1/(n+2)))=tan^(−1) ((((n+2)−(n+1))/((1+(n+1)(n+2)))) tan^(−1) ((n+2)−(n+1))= S_N =Σ_(n=0) ^(N=0) tan^(−1) (n+2)−tan^(−1) (n+1) n=0,1,2,3... S_N =tan^(−1) (N+2)−tan^(−1) (1) S=lim_(N→∞) [tan^(−1) (N+2)−tan^(−1) (1)] S=(π/2)−(π/4)=(π/4)](https://www.tinkutara.com/question/Q227057.png)
$${Rewritten}\:{using}\:{partial}\:{fractions} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}=\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)}{\left(\mathrm{1}+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\right.}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)\right)= \\ $$$${S}_{{N}} =\underset{{n}=\mathrm{0}} {\overset{{N}=\mathrm{0}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left({n}+\mathrm{2}\right)−\mathrm{tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right) \\ $$$${n}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}… \\ $$$${S}_{{N}} =\mathrm{tan}^{−\mathrm{1}} \left({N}+\mathrm{2}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${S}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{tan}^{−\mathrm{1}} \left({N}+\mathrm{2}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)\right] \\ $$$${S}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$