Menu Close

If-a-n-gt-0-and-lim-n-a-n-0-Find-lim-n-1-n-k-1-n-ln-k-n-a-n-

Tinku Tara 0 Comments
Pin




Question Number 209580 by hardmath last updated on 15/Jul/24
If a_n >0 and lim_(n→∞) a_n = 0 Find: lim_(n→∞) (1/n) Σ_(k=1) ^n ln ((k/n) + a_n ) = ?
$$\mathrm{If}\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} >\mathrm{0}\:\:\:\mathrm{and}\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:=\:\mathrm{0} \\ $$$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{ln}\:\left(\frac{\mathrm{k}}{\mathrm{n}}\:+\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} \right)\:=\:? \\ $$
Answered by mr W last updated on 15/Jul/24
=lim_(a→0) ∫_0 ^1 ln (x+a)dx =lim_(a→0) [ln (((1+a)^((1+a)) )/a^a )−1] =lim_(a→0) [ln (1+a)(1+(1/a))^a −1] =ln 1−1 =−1
$$=\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\:\left({x}+{a}\right){dx} \\ $$$$=\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\mathrm{ln}\:\frac{\left(\mathrm{1}+{a}\right)^{\left(\mathrm{1}+{a}\right)} }{{a}^{{a}} }−\mathrm{1}\right] \\ $$$$=\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\mathrm{ln}\:\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)^{{a}} −\mathrm{1}\right] \\ $$$$=\mathrm{ln}\:\mathrm{1}−\mathrm{1} \\ $$$$=−\mathrm{1} \\ $$
Commented by hardmath last updated on 16/Jul/24
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Berbere last updated on 16/Jul/24
∀ε≥0 ∃η>0 ∀n≥N a_n ≤ε; let n ≥N⇒ ⇒(1/n)Σ_(k=1) ^n ln((k/n))≤U_n =(1/n)Σ_(k=1) ^n ln((k/n)+a_n )≤(1/n)Σ_(k=1) ^n ln((k/n)+ε) ⇒(1/n)Σ_(k=1) ^n ln((k/n))≤U_n ≤∫_0 ^1 ln(x+ε)=(1+ε)ln(1+ε)−1 ∀ε>0 ∃N∈N such that U_n ≤(1+ε)ln(1+ε)−1 ⇒l=lim_(n→∞) U_n ≤inf (1+ε)ln(1+ε)−1=−1 ⇒∫_0 ^1 ln(x)dx≤l≤−1⇒l=−1
$$\forall\epsilon\geqslant\mathrm{0}\:\exists\eta>\mathrm{0}\:\forall{n}\geqslant\mathbb{N}\: \\ $$$${a}_{{n}} \leqslant\epsilon;\:{let}\:{n}\:\geqslant{N}\Rightarrow \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\frac{{k}}{{n}}\right)\leqslant{U}_{{n}} =\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\frac{{k}}{{n}}+{a}_{{n}} \right)\leqslant\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\frac{{k}}{{n}}+\epsilon\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\frac{{k}}{{n}}\right)\leqslant{U}_{{n}} \leqslant\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}+\epsilon\right)=\left(\mathrm{1}+\epsilon\right){ln}\left(\mathrm{1}+\epsilon\right)−\mathrm{1} \\ $$$$\forall\epsilon>\mathrm{0}\:\exists{N}\in\mathbb{N}\:{such}\:{that}\:{U}_{{n}} \leqslant\left(\mathrm{1}+\epsilon\right){ln}\left(\mathrm{1}+\epsilon\right)−\mathrm{1} \\ $$$$\Rightarrow{l}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{U}_{{n}} \leqslant{inf}\:\left(\mathrm{1}+\epsilon\right){ln}\left(\mathrm{1}+\epsilon\right)−\mathrm{1}=−\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){dx}\leqslant{l}\leqslant−\mathrm{1}\Rightarrow{l}=−\mathrm{1} \\ $$$$ \\ $$
Commented by hardmath last updated on 16/Jul/24
thank you dear professor cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *