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Question Number 148631 by tabata last updated on 29/Jul/21

find the region converge of the series and find the sum Σ_(n=1) ^∞ (n/(2^n (z−1)^n )) ?

$${find}\:{the}\:{region}\:{converge}\:{of}\:{the}\:{series}\:{and}\: \\ $$$${find}\:{the}\:{sum}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}}{\mathrm{2}^{{n}} \left({z}−\mathrm{1}\right)^{{n}} }\:? \\ $$

Commented by tabata last updated on 29/Jul/21

????

$$???? \\ $$

Answered by mathmax by abdo last updated on 30/Jul/21

u_n =(n/(2^n (z−1)^n )) ⇒∣(u_(n+1) /u_n )∣=∣((n+1)/(2^(n+1) (z−1)^(n+1) ))×((2^n (z−1)^n )/n)∣ =∣((n+1)/(2n(z−1)))∣→(1/(2∣z−1∣))<1 ⇒2∣z−1∣>1 ⇒∣z−1∣>(1/2) ⇒ donc la serie converge dans Λ={z ∈C /∣z−1∣>(1/2)}

$$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{n}}{\mathrm{2}^{\mathrm{n}} \left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{n}} }\:\Rightarrow\mid\frac{\mathrm{u}_{\mathrm{n}+\mathrm{1}} }{\mathrm{u}_{\mathrm{n}} }\mid=\mid\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }×\frac{\mathrm{2}^{\mathrm{n}} \left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\mid \\ $$$$=\mid\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}\left(\mathrm{z}−\mathrm{1}\right)}\mid\rightarrow\frac{\mathrm{1}}{\mathrm{2}\mid\mathrm{z}−\mathrm{1}\mid}<\mathrm{1}\:\Rightarrow\mathrm{2}\mid\mathrm{z}−\mathrm{1}\mid>\mathrm{1}\:\Rightarrow\mid\mathrm{z}−\mathrm{1}\mid>\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{donc}\:\mathrm{la}\:\mathrm{serie}\:\mathrm{converge}\:\mathrm{dans}\:\Lambda=\left\{\mathrm{z}\:\in\mathrm{C}\:/\mid\mathrm{z}−\mathrm{1}\mid>\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$ \\ $$

Commented by Sozan last updated on 30/Jul/21

sir and the sum how can find ?

$${sir}\:{and}\:{the}\:{sum}\:{how}\:{can}\:{find}\:? \\ $$

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