# li-n-m-k-1-n-8n-2-n-4-1-

Question Number 11036 by suci last updated on 08/Mar/17
$${l}\underset{{n}\rightarrow\sim} {{i}m}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{8}{n}^{\mathrm{2}} }{{n}^{\mathrm{4}} +\mathrm{1}}\:=….? \\$$
Commented by FilupS last updated on 08/Mar/17
$${S}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{8}{n}^{\mathrm{2}} }{{n}^{\mathrm{4}} +\mathrm{1}} \\$$$${S}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\infty}{\infty} \\$$$${S}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{16}{n}}{\mathrm{4}{n}^{\mathrm{3}} } \\$$$${S}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{16}}{\mathrm{4}{n}^{\mathrm{2}} } \\$$$${S}=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\$$$${n}\rightarrow\infty \\$$$${S}=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{0} \\$$$${S}=\mathrm{0} \\$$
Answered by ajfour last updated on 09/Mar/17
$$\mathrm{2}\pi\sqrt{\mathrm{2}}\:\:. \\$$
Commented by ajfour last updated on 09/Mar/17
$${reduces}\:{to}\:\:\:\mathrm{4}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \sqrt{\mathrm{tan}\:\theta\:}{d}\theta\:=\:\mathrm{2}\pi\sqrt{\mathrm{2}}\:. \\$$
Commented by FilupS last updated on 09/Mar/17
$$\mathrm{proof}? \\$$
Commented by ajfour last updated on 12/Mar/17
$${L}={lim}_{{n}\rightarrow\infty\:} \underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{8}{k}^{\mathrm{2}} }{{k}^{\mathrm{4}} +\mathrm{1}}\right) \\$$$${L}={lim}_{{n}\rightarrow\infty\:} {S}_{{n}} \\$$$${S}_{{n}} =\frac{\mathrm{8}}{{n}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\left({k}/{n}\right)^{\mathrm{2}} }{\left({k}/{n}\right)^{\mathrm{4}} +\left(\mathrm{1}/{n}\right)^{\mathrm{4}} }\:\left(\frac{\mathrm{1}}{{n}}\right) \\$$$${now}\:\frac{{k}}{{n}}\:={x}\:;\:\frac{\mathrm{1}}{{n}}=\frac{\left({k}+\mathrm{1}\right)−{k}}{{n}}\:={dx} \\$$$${S}_{{n}} =\:\frac{\mathrm{8}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +{a}^{\mathrm{4}} }\:{dx}\: \\$$$${where}\:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:={a}^{\mathrm{4}} \:\left({for}\:{ease}\right) \\$$$${let}\:{x}^{\mathrm{2}} ={a}^{\mathrm{2}} \mathrm{tan}\:\theta\Rightarrow{x}={a}\sqrt{\mathrm{tan}\:\theta} \\$$$$\Rightarrow\:{x}^{\mathrm{4}} +{a}^{\mathrm{4}} ={a}^{\mathrm{4}} \mathrm{sec}\:^{\mathrm{2}} \theta \\$$$${dx}=\frac{{a}\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta}{\mathrm{2}\sqrt{\mathrm{tan}\:\theta}}\:\:,\:{substituting} \\$$$${S}_{{n}} =\frac{\mathrm{8}}{{n}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\:\frac{\left({a}^{\mathrm{3}} /\mathrm{2}\right)\mathrm{sec}\:^{\mathrm{2}} \theta\sqrt{\mathrm{tan}\:\theta}{d}\theta}{{a}^{\mathrm{4}} \mathrm{sec}\:^{\mathrm{2}} \theta} \\$$$$\:\:{L}={lim}_{{n}\rightarrow\infty\:\:} \left[\:\frac{\mathrm{8}}{\mathrm{2}{na}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\sqrt{\mathrm{tan}\:\theta}\:{d}\theta\right] \\$$$$\:\:{as}\:{n}\rightarrow\:\infty,\:{na}\rightarrow\mathrm{1}\:{for}\:{a}^{\mathrm{4}} =\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\$$$${so}\:{L}=\:\mathrm{4}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\sqrt{\mathrm{tan}\:\theta}\:{d}\theta \\$$$${L}=\mathrm{4}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{2}\pi\sqrt{\mathrm{2}}\:\:. \\$$
Commented by FilupS last updated on 12/Mar/17
$$\mathrm{oh},\:\mathrm{they}\:\mathrm{made}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\$$$$\mathrm{i}\:\mathrm{see}\:\mathrm{now} \\$$
Commented by ajfour last updated on 12/Mar/17
$${the}\:\Sigma\:{should}\:{make}\:{sense}. \\$$
Answered by DrDaveR last updated on 22/Mar/17
$$−\mathrm{4}\pi.{Imag}\left(\left(\sqrt{−{i}}\right){cot}\left(\pi\left(\sqrt{−{i}}\right)\right)\:{which}\:{is}\:{about}\:\mathrm{9}.\mathrm{028}.\:{Rewrite}\:{the}\right. \\$$$${function}\:{in}\:{the}\:{sum}\:{as}\:\mathrm{4}\left(\mathrm{1}/\left({k}^{\mathrm{2}} +{i}\right)+\mathrm{1}/\left({k}^{\mathrm{2}} −{i}\right)\right)\:{then}\:{use}\: \\$$$${contangent}\:{formula}\:{for}\:{the}\:{sum}\:{of}\:{these}\:{series}\:{instead} \\$$