Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 130845 by mnjuly1970 last updated on 29/Jan/21

           ...  nice  calculus...    calculate::     lim_( n→∞) Σ_(k=1) ^n (k/(2n^2 +k)) =?

$$\:\:\:\:\:\:\:\:\:\:\:...\:\:{nice}\:\:{calculus}... \\ $$$$\:\:{calculate}:: \\ $$$$\:\:\:{lim}_{\:{n}\rightarrow\infty} \underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} +{k}}\:=? \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 29/Jan/21

1≤k≤n ⇒2n^2 +1≤2n^2 +k≤2n^2 +n  ⇒(k/(2n^2 +n))≤(k/(2n^2 +k))≤(k/(2n^2 +1))  ⇒((n(n+1))/(2(2n^2 +n)))≤Σ_(k=1) ^n (k/(2n^2 +k))≤((n(n+1))/(2(2n^2 +1)))  ⇒(1/4)≤lim_(n→∞) Σ_(k=1) ^n (k/(2n^2 +k))≤(1/4)

$$\mathrm{1}\leqslant\mathrm{k}\leqslant\mathrm{n}\:\Rightarrow\mathrm{2n}^{\mathrm{2}} +\mathrm{1}\leqslant\mathrm{2n}^{\mathrm{2}} +\mathrm{k}\leqslant\mathrm{2n}^{\mathrm{2}} +\mathrm{n} \\ $$$$\Rightarrow\frac{\mathrm{k}}{\mathrm{2n}^{\mathrm{2}} +\mathrm{n}}\leqslant\frac{\mathrm{k}}{\mathrm{2n}^{\mathrm{2}} +\mathrm{k}}\leqslant\frac{\mathrm{k}}{\mathrm{2n}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{2n}^{\mathrm{2}} +\mathrm{n}\right)}\leqslant\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{k}}{\mathrm{2n}^{\mathrm{2}} +\mathrm{k}}\leqslant\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{2n}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\leqslant\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{k}}{\mathrm{2n}^{\mathrm{2}} +\mathrm{k}}\leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 29/Jan/21

hi master  brandon   answer is  (1/4)

$${hi}\:{master}\:\:{brandon}\:\:\:{answer}\:{is}\:\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 29/Jan/21

   in third line   ((n(n+1))/(2(2n^2 +n)))≤A_n ≤((n(n+1))/(2(2n^2 +1)))    your  answer  is correct   lim_(n→∞) A_n =(1/4)

$$\:\:\:{in}\:{third}\:{line}\:\:\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{2}{n}^{\mathrm{2}} +{n}\right)}\leqslant{A}_{{n}} \leqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}\right)}\:\: \\ $$$${your}\:\:{answer}\:\:{is}\:{correct} \\ $$$$\:{lim}_{{n}\rightarrow\infty} {A}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by Ar Brandon last updated on 29/Jan/21

Oh! thanks for remark, Sir  😃

$$\mathrm{Oh}!\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{remark},\:\mathrm{Sir} \\ $$😃

Terms of Service

Privacy Policy

Contact: info@tinkutara.com