Question Number 57488 by Abdo msup. last updated on 05/Apr/19
![let A_n =∫_0 ^n ((t[t])/(3+t^2 ))dt 1)calculate lim_(n→+∞) A_n 2) find nature if Σ A_n](https://www.tinkutara.com/question/Q57488.png)
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{{t}\left[{t}\right]}{\mathrm{3}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right){calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{if}\:\Sigma\:{A}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 06/Apr/19
![1) we have A_n =Σ_(k=0) ^(n−1) ∫_k ^(k+1) ((tk)/(3+t^2 )) dt =Σ_(k=0) ^(n−1) k ∫_k ^(k+1) ((tdt)/(t^2 +3)) =(1/2)Σ_(k=0) ^(n−1) k [ ln(t^2 +3)]_k ^(k+1) =(1/2) Σ_(k=0) ^(n−1) k ln{(((k+1)^2 +3)/(k^2 +3))} =(1/2) Σ_(k=0) ^(n−1) k ln{((k^2 +3 +2k)/(k^2 +3))} =(1/2) Σ_(k=1) ^(n−1) k ln{ 1+((2k)/(k^2 +3))} . =(1/2){ln(1+(2/4)) +2 ln( 1+(4/7)) +3 ln(1+(6/(12)))+....+(n−1)ln(1+((2n−2)/((n−1)^2 +3)))}](https://www.tinkutara.com/question/Q57527.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\:\frac{{tk}}{\mathrm{3}+{t}^{\mathrm{2}} }\:{dt}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{tdt}}{{t}^{\mathrm{2}} \:+\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:\left[\:{ln}\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)\right]_{{k}} ^{{k}+\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:{ln}\left\{\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}}{{k}^{\mathrm{2}} \:+\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:{ln}\left\{\frac{{k}^{\mathrm{2}} \:+\mathrm{3}\:+\mathrm{2}{k}}{{k}^{\mathrm{2}} \:+\mathrm{3}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{k}\:{ln}\left\{\:\mathrm{1}+\frac{\mathrm{2}{k}}{{k}^{\mathrm{2}} \:+\mathrm{3}}\right\}\:. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{4}}\right)\:+\mathrm{2}\:{ln}\left(\:\mathrm{1}+\frac{\mathrm{4}}{\mathrm{7}}\right)\:+\mathrm{3}\:{ln}\left(\mathrm{1}+\frac{\mathrm{6}}{\mathrm{12}}\right)+….+\left({n}−\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{2}{n}−\mathrm{2}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}}\right)\right\} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Apr/19
![∫_0 ^n ((t[t])/(3+t^2 ))dt (1/2)∫_0 ^1 ((d(3+t^2 ))/(3+t^2 ))×0+(1/2)∫_1 ^2 ((d(3+t^2 ))/(3+t^2 ))×1+(1/2)∫_2 ^3 ((d(3+t^2 ))/(3+t^2 ))×2dt+...+(1/2)∫_(n−1) ^n ((d(3+t^2 ))/(3+t^2 ))×(n−1)dt =(1/2)[0×∣ln(3+t^2 )∣_0 ^1 +1×∣ln(3+t^2 )∣_1 ^2 +2×∣ln(3+t^2 )∣_2 ^3 +..+(n−1)∣ln(3+t^2 )∣_(n−1) ^n ] =(1/2)Σ_(n=1) ^n (n−1)ln{((3+n^2 )/(3+(n−1)^2 ))}](https://www.tinkutara.com/question/Q57508.png)
$$\int_{\mathrm{0}} ^{{n}} \frac{{t}\left[{t}\right]}{\mathrm{3}+{t}^{\mathrm{2}} }{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)}{\mathrm{3}+{t}^{\mathrm{2}} }×\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{d}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)}{\mathrm{3}+{t}^{\mathrm{2}} }×\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{d}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)}{\mathrm{3}+{t}^{\mathrm{2}} }×\mathrm{2}{dt}+…+\frac{\mathrm{1}}{\mathrm{2}}\int_{{n}−\mathrm{1}} ^{{n}} \frac{{d}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)}{\mathrm{3}+{t}^{\mathrm{2}} }×\left({n}−\mathrm{1}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{0}×\mid{ln}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{1}×\mid{ln}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)\mid_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}×\mid{ln}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)\mid_{\mathrm{2}} ^{\mathrm{3}} +..+\left({n}−\mathrm{1}\right)\mid{ln}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)\mid_{{n}−\mathrm{1}} ^{{n}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}−\mathrm{1}\right){ln}\left\{\frac{\mathrm{3}+{n}^{\mathrm{2}} }{\mathrm{3}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$