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Question Number 181644 by mathlove last updated on 28/Nov/22

$$provethat$$$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{1+\frac{1}{3}+\frac{1}{5}+\cdot \cdot \cdot \cdot +\frac{1}{2x+1}}{xln\sqrt{x}}\right)}^{ln\sqrt{x}}=2e^{\frac{\gamma }{2}}$$$$$$

Answered by aleks041103 last updated on 28/Nov/22

Answered by aleks041103 last updated on 28/Nov/22

$$\underset{k=1}{\sum ^{x+1}}\frac{1}{2k-1}=\underset{k=1}{\sum ^{x+1}}(\frac{1}{2k-1}+\frac{1}{2k})-\underset{k=1}{\sum ^{x+1}}\frac{1}{2k}=$$$$=H(2x+2)-\frac{1}{2}H(x+1)=$$$$=H(2x+1)+\frac{1}{2x+2}-\frac{1}{2}H\left(x\right)+\frac{1}{2(x+1)}=$$$$=H(2x+1)-\frac{1}{2}H\left(x\right)$$$$H\left(x\right)\to ln\left(x\right)+\gamma$$$$\Rightarrow \underset{k=1}{\sum ^{x+1}}\frac{1}{2k-1}\to ln(2x+1)-\frac{1}{2}ln\left(x\right)+\frac{\gamma }{2}$$$$\Rightarrow L=\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{ln(2{\left(\sqrt{x}\right)}^2+1)-ln\left(\sqrt{x}\right)+\frac{\gamma }{2}}{ln\left(\sqrt{x}\right)}\right)}^{ln\left(\sqrt{x}\right)}=$$$$=\underset{y\to \mathrm{\infty }}{lim}{\left(\frac{ln(2y^2+1)-ln\left(y\right)+\gamma /2}{ln\left(y\right)}\right)}^{ln\left(y\right)}=$$$$=\underset{y\to \mathrm{\infty }}{lim}{(1+\frac{ln(2y^2+1)-ln\left(y^2\right)+\gamma /2}{ln\left(y\right)})}^{ln\left(y\right)}=$$$$=\underset{y\to \mathrm{\infty }}{lim}{(1+\frac{ln(2+y^{-2})+\gamma /2}{ln\left(y\right)})}^{ln\left(y\right)}=$$$$=exp\left(\underset{y\to \mathrm{\infty }}{lim}(ln(2+y^{-2})+\gamma /2)\right)=$$$$=e^{\frac{\gamma }{2}+ln\left(2\right)}=2e^{\gamma /2}$$

Commented by aleks041103 last updated on 28/Nov/22

Commented by mathlove last updated on 28/Nov/22

$$thanks$$

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