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Question Number 40123 by maxmathsup by imad last updated on 15/Jul/18
$${study}\:{the}\:{convergence}\:{of} \\ $$$${v}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{\mathrm{2}^{{k}} \:+{ln}\left({k}\right)} \\ $$
Answered by math khazana by abdo last updated on 19/Jul/18
![we have v_n =(1/2) +Σ_(k=2) ^n (((−1)^(k+1) )/(2^k +ln(k))) ⇒ ∣v_n ∣ ≤ (1/2) +Σ_(k=2) ^n (1/(2^k +ln(k))) but for k∈[[2,n]] ln(k)≥ln(2)>0 ⇒2^k +ln(k)>2^k ⇒ (1/(2^k +ln(k))) < (1/2^k ) ⇒∣v_n ∣ <(1/2) +Σ_(k=2) ^n (1/2^k ) =Σ_(k=1) ^n (1/2^k ) the serie Σ_(k≥1) (1/2^k ) is convergent so (v_n ) converges](Q40307.png)
$${we}\:{have}\:{v}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:+\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{\mathrm{2}^{{k}} \:+{ln}\left({k}\right)}\:\Rightarrow \\ $$$$\mid{v}_{{n}} \mid\:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}}\:+\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} \:+{ln}\left({k}\right)}\:{but}\:{for}\:{k}\in\left[\left[\mathrm{2},{n}\right]\right] \\ $$$${ln}\left({k}\right)\geqslant{ln}\left(\mathrm{2}\right)>\mathrm{0}\:\Rightarrow\mathrm{2}^{{k}} \:+{ln}\left({k}\right)>\mathrm{2}^{{k}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{k}} \:+{ln}\left({k}\right)}\:<\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\Rightarrow\mid{v}_{{n}} \mid\:<\frac{\mathrm{1}}{\mathrm{2}}\:+\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}^{{k}} } \\ $$$${the}\:{serie}\:\sum_{{k}\geqslant\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:{is}\:{convergent}\:{so}\:\left({v}_{{n}} \right)\:{converges} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$