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Question Number 114625 by bobhans last updated on 20/Sep/20

Answered by john santu last updated on 20/Sep/20

⇔k^4 +2k^2 +9=k^4 +6k^2 +9−4k^2 = (k^2 −2k+3)(k^2 +2k+3) S=Σ_(k=1) ^∞ (k/((k^2 −2k+3)(k^2 +2k+3))) S=Σ_(k=1) ^∞ (1/4)((1/(k^2 −2k+3)) − (1/(k^2 +2k+3))) S= (1/4)Σ_(k=1) ^∞ ((1/(k(k−2)+3)) − (1/(k(k+2)+3))) S=(1/4)((1/2)+(1/3)−0)=(5/(24)).

$$\Leftrightarrow{k}^{\mathrm{4}} +\mathrm{2}{k}^{\mathrm{2}} +\mathrm{9}={k}^{\mathrm{4}} +\mathrm{6}{k}^{\mathrm{2}} +\mathrm{9}−\mathrm{4}{k}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{3}\right)\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{3}\right) \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}}{\left({k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{3}\right)\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{3}\right)} \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{3}}\:−\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{3}}\right) \\ $$$${S}=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}\left({k}−\mathrm{2}\right)+\mathrm{3}}\:−\:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{2}\right)+\mathrm{3}}\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{0}\right)=\frac{\mathrm{5}}{\mathrm{24}}. \\ $$

Commented by bobhans last updated on 20/Sep/20

gives “like“

$${gives}\:``{like}`` \\ $$

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