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Question Number 196471 by Tawa11 last updated on 25/Aug/23

lim_(n→∞) (1/n)((1/(2 + (1/n))) + (1/(2 + (2/n))) + (1/(2 + (3/n))) + .. + (1/(2 + (n/n))))

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}}\left(\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{n}}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{2}}{\mathrm{n}}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{3}}{\mathrm{n}}}\:\:+\:\:..\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{n}}{\mathrm{n}}}\right) \\ $$

Answered by mr W last updated on 25/Aug/23

=∫_0 ^1 (dx/(2+x)) =[ln (2+x)]_0 ^1 =ln 3−ln 2 =ln (3/2)

$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{2}+{x}} \\ $$$$=\left[\mathrm{ln}\:\left(\mathrm{2}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{2} \\ $$$$=\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 25/Aug/23

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by Tawa11 last updated on 25/Aug/23

Sir is the sum a general term I should know? Or you solved to get ∫_0 ^1 (1/(2 + x)) dx

$$\mathrm{Sir}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{a}\:\mathrm{general}\:\mathrm{term}\:\mathrm{I}\:\mathrm{should}\:\mathrm{know}? \\ $$$$\mathrm{Or}\:\mathrm{you}\:\mathrm{solved}\:\mathrm{to}\:\mathrm{get}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}\:+\:\mathrm{x}}\:\mathrm{dx} \\ $$

Commented by mr W last updated on 25/Aug/23

∫_0 ^1 (1/(2+x))dx =lim_(Δx→0) Σ(1/(2+x_k ))×Δx =lim_(n→∞) Σ_(k=1) ^n (1/(2+(k/n)))×(1/n) =lim_(n→∞) ((1/n)Σ_(k=1) ^n (1/(2+(k/n))))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}+{x}}{dx} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\Sigma\frac{\mathrm{1}}{\mathrm{2}+{x}_{{k}} }×\Delta{x} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}+\frac{{k}}{{n}}}×\frac{\mathrm{1}}{{n}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}+\frac{{k}}{{n}}}\right) \\ $$

Commented by mr W last updated on 25/Aug/23

Commented by mr W last updated on 25/Aug/23

generally Δx=(1/n) ∫_0 ^1 f(x)dx≈y_1 Δx+y_2 Δx+...+y_k Δx+...+y_n Δx ∫_0 ^1 f(x)dx≈(y_1 +y_2 +...+y_k +...+y_n )Δx ∫_0 ^1 f(x)dx≈[f(x_1 )+f(x_2 )+...+f(x_k )+...+f(x_n )]Δx ∫_0 ^1 f(x)dx≈[f(Δx)+f(2Δx)+...+f(kΔx)+...+f(nΔx)]Δx ∫_0 ^1 f(x)dx≈[f((1/n))+f((2/b))+...+f((k/n))+...+f((n/n))](1/n) ∫_0 ^1 f(x)dx≈(1/n)Σ_(k=1) ^n f((k/n)) ∫_0 ^1 f(x)dx=lim_(n→∞) [(1/n)Σ_(k=1) ^n f((k/n))] in current case: f(x)=(1/(2+x))

$${generally} \\ $$$$\Delta{x}=\frac{\mathrm{1}}{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\approx{y}_{\mathrm{1}} \Delta{x}+{y}_{\mathrm{2}} \Delta{x}+...+{y}_{{k}} \Delta{x}+...+{y}_{{n}} \Delta{x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\approx\left({y}_{\mathrm{1}} +{y}_{\mathrm{2}} +...+{y}_{{k}} +...+{y}_{{n}} \right)\Delta{x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\approx\left[{f}\left({x}_{\mathrm{1}} \right)+{f}\left({x}_{\mathrm{2}} \right)+...+{f}\left({x}_{{k}} \right)+...+{f}\left({x}_{{n}} \right)\right]\Delta{x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\approx\left[{f}\left(\Delta{x}\right)+{f}\left(\mathrm{2}\Delta{x}\right)+...+{f}\left({k}\Delta{x}\right)+...+{f}\left({n}\Delta{x}\right)\right]\Delta{x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\approx\left[{f}\left(\frac{\mathrm{1}}{{n}}\right)+{f}\left(\frac{\mathrm{2}}{{b}}\right)+...+{f}\left(\frac{{k}}{{n}}\right)+...+{f}\left(\frac{{n}}{{n}}\right)\right]\frac{\mathrm{1}}{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\approx\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left(\frac{{k}}{{n}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left(\frac{{k}}{{n}}\right)\right] \\ $$$${in}\:{current}\:{case}:\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}+{x}} \\ $$

Commented by Tawa11 last updated on 25/Aug/23

Wow, I really appreciate sir.

$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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