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Question Number 142667 by mnjuly1970 last updated on 03/Jun/21

Answered by mindispower last updated on 03/Jun/21

$$T\left(n\right)={\int }_0^1{x}^{n}ln(1+x)dx=\frac{ln\left(2\right)}{n+1}-{\int }_0^1\frac{x^{n+1}}{n+1}.\frac{dx}{1+x}$$$$0⩽{\int }_0^1\frac{x^{n+1}}{1+x}dx⩽{\int }_0^1{x}^{n+1}dx=\frac{1}{n+2}$$$$\frac{ln\left(2\right)}{n+1}⩾T\left(n\right)⩾\frac{ln\left(2\right)}{n+1}-\frac{1}{(n+1)(n+2)}$$$$\Rightarrow \frac{n+2}{n+1}ln\left(2\right)-\frac{1}{n+1}⩽(n+2)T\left(n\right)⩽\frac{n+2}{n+1}ln\left(2\right)$$$$\underset{n\to \mathrm{\infty }}{\lim }T\left(n\right)=ln\left(2\right)$$$$$$$$$$

Commented by mnjuly1970 last updated on 03/Jun/21

$$thanksalot....$$

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