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Question Number 211421 by universe last updated on 08/Sep/24

$$\mathrm{if}{\mathrm{a}}_{\mathrm{n}}={\mathrm{n}}^4{\int }_{\mathrm{n}}^{\mathrm{n}+1}\frac{\mathrm{x}\mathrm{dx}}{1+{\mathrm{x}}^5}\mathrm{then}$$$$\left(1\right)\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{n}}\mathrm{is}\mathrm{convergent}\mathrm{or}\mathrm{divergent}??$$$$\left(2\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\mathrm{a}}_{\mathrm{n}}=??$$

Answered by mathmax last updated on 09/Sep/24

$$n⩽x⩽n+1et1+n^5⩽1+x^5⩽1+{(n+1)}^5\Rightarrow$$$$\frac{1}{1+{(n+1)}^5}⩽\frac{1}{1+x^5}⩽\frac{1}{1+n^5}\Rightarrow \frac{n^4\times n}{{(n+1)}^5+1}⩽n^4\frac{x}{1+x^5}⩽\frac{n^4(n+1)}{1+n^5}$$$$\Rightarrow \frac{n^5}{{(n+1)}^5+1}⩽n^4{\int }_n^{n+1}\frac{xdx}{1+x^5}⩽\frac{n^5+n}{n^5+1}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{a}_n=1et\mathrm{\Sigma }{a}_{n}estdiv.$$

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