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Question Number 151660 by mathdanisur last updated on 22/Aug/21

Answered by Olaf_Thorendsen last updated on 22/Aug/21

$$\ln (e+\sin kx)=1+\ln (1+\frac{\sin kx}{e})\underset{0}{\sim }1+\frac{kx}{e}$$$$\frac{1-\underset{k=1}{\prod ^n}\ln (e+\sin kx)}{x}\underset{0}{\sim }\frac{1-\underset{k=1}{\prod ^n}(1+\frac{kx}{e})}{x}\left(1\right)$$$$\mathrm{Let}{\mathrm{P}}_n\left(x\right)=\underset{k=1}{\prod ^n}(1+\frac{kx}{e})$$$${\mathrm{lnP}}_n\left(x\right)=\underset{k=1}{\sum ^n}\ln (1+\frac{kx}{e})$$$${\mathrm{lnP}}_n\left(x\right)\underset{0}{\sim }\underset{k=1}{\sum ^n}\frac{kx}{e}=\frac{n(n+1)}{2e}x$$$$\left(1\right)\underset{0}{\sim }\frac{1-e^{\frac{n(n+1)}{2e}x}}{x}\underset{0}{\sim }\frac{1-(1+\frac{n(n+1)}{2e}x)}{x}$$$$\underset{0}{\sim }-\frac{n(n+1)}{2e}$$

Commented by mathdanisur last updated on 23/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}$$

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