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Question Number 87301 by M±th+et£s last updated on 03/Apr/20

Commented by mathmax by abdo last updated on 04/Apr/20

$$letI={\int }_1^{+\mathrm{\infty }}\frac{dx}{2{\left[x\right]}^2+\left[x\right]}\Rightarrow I=\sum _{n=1}^{\mathrm{\infty }}{\int }_n^{n+1}\frac{dx}{2n^2+n}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(2n+1)}={lim}_{n\to +\mathrm{\infty }}\sum _{k=1}^n\frac{1}{k(2k+1)}={lim}_{n\to +\mathrm{\infty }}{S}_n$$$$\frac{1}{2}{S}_n=\sum _{k=1}^n(\frac{1}{2k}-\frac{1}{2k+1})=\frac{1}{2}\sum _{k=1}^n\frac{1}{k}-\sum _{k=1}^n\frac{1}{2k+1}$$$$\sum _{k=1}^n\frac{1}{k}=H_n$$$$\sum _{k=1}^n\frac{1}{2k+1}=\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n}+\frac{1}{2n+1}$$$$-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2n}=H_{2n+1}-1-\frac{1}{2}{H}_n\Rightarrow$$$$\frac{1}{2}{S}_n=\frac{1}{2}{H}_n-H_{2n+1}+\frac{1}{2}{H}_n+1$$$$=(ln\left(n\right)+\gamma +o\left(\frac{1}{n}\right))-ln(2n+1)-\gamma +o\left(\frac{1}{2n+1}\right)+1$$$$=ln\left(\frac{n}{2n+1}\right)+1\Rightarrow 1-ln\left(2\right)\Rightarrow S_n\to 2-2ln\left(2\right)(n\to +\mathrm{\infty })$$$$wehaveln\left(\frac{e^2}{4}\right)=2ln\left(e\right)-2ln\left(2\right)=2-2ln\left(2\right)$$$$I=lim{S}_n=2-2ln\left(2\right)=ln\left(\frac{e^2}{4}\right).$$

Commented by M±th+et£s last updated on 04/Apr/20

$$godblessyousir.greatsolution$$

Commented by abdomathmax last updated on 04/Apr/20

$$youarewelcome.$$

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