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Question Number 32303 by abdo imad last updated on 22/Mar/18
$${find}\:{lim}_{{n}\rightarrow\infty} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{n}+{k}}\:\:. \\ $$
Commented by abdo imad last updated on 01/Apr/18
![let put S_n = Σ_(k=1) ^n (1/(2n+k)) S_(n ) =(1/n) Σ_(k=1) ^n (1/(2 +(k/n))) ⇒ S_n is a Rieman sum and lim_(n→∞) S_n = ∫_0 ^1 (dx/(2+x)) =[ln∣2+x∣]_0 ^1 = ln3 −ln2 .](Q32766.png)
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{n}+{k}} \\ $$$${S}_{{n}\:} \:\:=\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}\:+\frac{{k}}{{n}}}\:\Rightarrow\:{S}_{{n}} \:{is}\:{a}\:{Rieman}\:{sum}\:{and} \\ $$$${lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{2}+{x}}\:\:=\left[{ln}\mid\mathrm{2}+{x}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:{ln}\mathrm{3}\:−{ln}\mathrm{2}\:. \\ $$