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Question Number 50293 by pooja24 last updated on 15/Dec/18

lim_(x→0)  Σ_(r=1) ^(2n) ((1/(r+n)))=?  please help

$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left(\frac{\mathrm{1}}{{r}+{n}}\right)=? \\ $$$${please}\:{help} \\ $$

Commented by maxmathsup by imad last updated on 15/Dec/18

let S_n =Σ_(k=1) ^(2n)  (1/(k+n)) ⇒S_n =Σ_(k=1) ^n  (1/(k+n)) +Σ_(k=n+1) ^(2n)  (1/(k+n)) =A_n  +B_n   we have A_n =(1/n) Σ_(k=1) ^n   (1/(1+(k/n))) →_(n→+∞)   ∫_0 ^1   (dx/(1+x)) =ln∣1+x∣]_0 ^1 =ln(2)  changement k−n =p  give B_n =Σ_(p=1) ^n   (1/(n+p+n)) =Σ_(p=1) ^n   (1/(p+2n))  =(1/n) Σ_(p=1) ^n   (1/(2+(p/n))) →_(n→+∞)   ∫_0 ^1    (dx/(2+x)) =[ln∣2+x∣]_0 ^1  =ln(x)−ln(2) ⇒  lim_(n→+∞)  S_n  =lim_(n→+∞ ) (A_n +B_n )=ln(2)+ln(3)−ln(2) =ln(3)

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}+{n}}\:\Rightarrow{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+{n}}\:+\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}+{n}}\:={A}_{{n}} \:+{B}_{{n}} \\ $$$$\left.{we}\:{have}\:{A}_{{n}} =\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}}{{n}}}\:\rightarrow_{{n}\rightarrow+\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}}\:={ln}\mid\mathrm{1}+{x}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} ={ln}\left(\mathrm{2}\right) \\ $$$${changement}\:{k}−{n}\:={p}\:\:{give}\:{B}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{n}+{p}+{n}}\:=\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{p}+\mathrm{2}{n}} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}+\frac{{p}}{{n}}}\:\rightarrow_{{n}\rightarrow+\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{2}+{x}}\:=\left[{ln}\mid\mathrm{2}+{x}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:={ln}\left({x}\right)−{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:={lim}_{{n}\rightarrow+\infty\:} \left({A}_{{n}} +{B}_{{n}} \right)={ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)\:={ln}\left(\mathrm{3}\right) \\ $$

Commented by maxmathsup by imad last updated on 15/Dec/18

∫_0 ^1   (dx/(2+x)) =ln(3)−ln(2).

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

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

there is no ∞ in question...

$${there}\:{is}\:{no}\:\infty\:{in}\:{question}... \\ $$

Commented by maxmathsup by imad last updated on 15/Dec/18

i think the Queston is find lim_(n→+∞) Σ_(r=1) ^(2n)  (1/(r+n))

$${i}\:{think}\:{the}\:{Queston}\:{is}\:{find}\:{lim}_{{n}\rightarrow+\infty} \sum_{{r}=\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{r}+{n}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

T_r =(1/(r+n))  T_1 =(1/(1+n))  T_2 =(1/(2+n))  ..  ...  T_(2n) =(1/(2n+n))  S_(2n) =(1/(1+n))+(1/(2+n))+(1/(3+n))+...+(1/(2n+n))  now   (1/(1+n))<(1/n)  (1/(2+n))<(1/n)  ...  ...  (1/(2n+n))<(1/n)  so S_(2n) <2n×(1/n)  S_(2n) <2  in the problem no mention of x, so pls check  the question...  S_(2n) =Σ_(r=1) ^(2n) (1/(r+n))<2                    lim_(x→0)

$${T}_{{r}} =\frac{\mathrm{1}}{{r}+{n}} \\ $$$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}+{n}} \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}+{n}} \\ $$$$.. \\ $$$$... \\ $$$${T}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{2}{n}+{n}} \\ $$$${S}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{1}+{n}}+\frac{\mathrm{1}}{\mathrm{2}+{n}}+\frac{\mathrm{1}}{\mathrm{3}+{n}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}+{n}} \\ $$$${now}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{n}}<\frac{\mathrm{1}}{{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+{n}}<\frac{\mathrm{1}}{{n}} \\ $$$$... \\ $$$$... \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{n}+{n}}<\frac{\mathrm{1}}{{n}} \\ $$$${so}\:{S}_{\mathrm{2}{n}} <\mathrm{2}{n}×\frac{\mathrm{1}}{{n}} \\ $$$${S}_{\mathrm{2}{n}} <\mathrm{2} \\ $$$${in}\:{the}\:{problem}\:{no}\:{mention}\:{of}\:{x},\:{so}\:{pls}\:{check} \\ $$$${the}\:{question}... \\ $$$${S}_{\mathrm{2}{n}} =\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{r}+{n}}<\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}} \\ $$$$ \\ $$

Answered by ajfour last updated on 15/Dec/18

lim_(n→∞)  Σ_(r=1) ^(2n)  (1/(r+n)) = ∫_0 ^(  2) (dx/(x+1)) = ln 3 .

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\:\frac{\mathrm{1}}{{r}+{n}}\:=\:\int_{\mathrm{0}} ^{\:\:\mathrm{2}} \frac{{dx}}{{x}+\mathrm{1}}\:=\:\mathrm{ln}\:\mathrm{3}\:. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

there is no ∞ in question...

$${there}\:{is}\:{no}\:\infty\:{in}\:{question}... \\ $$

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