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Question Number 108705 by mathmax by abdo last updated on 18/Aug/20

$$\mathrm{if}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{k}}=\mathrm{n}(2^{\mathrm{n}}+3)\mathrm{determine}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{\mathrm{u}}_{\mathrm{k}}}$$

Answered by mathmax by abdo last updated on 21/Aug/20

$$\mathrm{we}\mathrm{have}{\mathrm{u}}_1+{\mathrm{u}}_2+....+{\mathrm{u}}_{\mathrm{n}}=\mathrm{n}(2^{\mathrm{n}}+3)$$$${\mathrm{u}}_1+{\mathrm{u}}_2+....+{\mathrm{u}}_{\mathrm{n}}+{\mathrm{u}}_{\mathrm{n}+1}=(\mathrm{n}+1)(2^{\mathrm{n}+1}+3)\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}+1}=(\mathrm{n}+1)2^{\mathrm{n}+1}+3\mathrm{n}+3-{\mathrm{n}2}^{\mathrm{n}}-3\mathrm{n}$$$$=2\mathrm{n}{2}^{\mathrm{n}}+2^{\mathrm{n}+1}-{\mathrm{n}2}^{\mathrm{n}}+3=\mathrm{n}{2}^{\mathrm{n}}+2.2^{\mathrm{n}}+3=(\mathrm{n}+2)2^{\mathrm{n}}+3\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}=(\mathrm{n}+1)2^{\mathrm{n}-1}+3\Rightarrow {\mathrm{v}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{\mathrm{u}}_{\mathrm{k}}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{(\mathrm{k}+1)2^{\mathrm{k}-1}+3}$$$$1⩽\mathrm{k}⩽\mathrm{n}\Rightarrow 2⩽\mathrm{k}+1⩽\mathrm{n}+1\mathrm{and}1⩽2^{\mathrm{k}-1}⩽2^{\mathrm{n}-1}\Rightarrow 2⩽(\mathrm{k}+1)2^{\mathrm{k}-1}⩽\mathrm{n}+1+2^{\mathrm{n}-1}\Rightarrow$$$$5⩽)\mathrm{k}+1)2^{\mathrm{k}-1}+3⩽\mathrm{n}+4+2^{\mathrm{n}-1}\Rightarrow \frac{1}{\mathrm{n}+4+2^{\mathrm{n}-1}}⩽\frac{1}{(...)}⩽\frac{1}{5}$$$$\frac{1}{\mathrm{n}+4+2^{\mathrm{n}-1}}=\frac{1}{2^{\mathrm{n}-1}\{\frac{\mathrm{n}+4}{2^{\mathrm{n}-1}}+1\}}\sim \frac{1}{2^{\mathrm{n}-1}}(1-\frac{\mathrm{n}+4}{2^{\mathrm{n}-1}})=\frac{1}{2^{\mathrm{n}-1}}-\frac{\mathrm{n}+4}{4^{\mathrm{n}-1}}$$$$\Rightarrow \mathrm{\Sigma }\frac{1}{\mathrm{n}+4+2^{\mathrm{n}-1}}\mathrm{converges}\Rightarrow \mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}\mathrm{cv}\mathrm{rest}\mathrm{to}\mathrm{find}\mathrm{limit}....$$$$\mathrm{be}\mathrm{continued}....$$

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