Question

Se rezolva cu ajutorul factorului comun fortat n lnn?

Answer 1

Nu stiu daca ai calculat deja integrala, dar o las aici si pe ea:

$\int_{0}^{k} \cfrac{1}{x^2+3x+2} \,dx=\int_{0}^{k} \cfrac{1}{x^2+3x+(\frac{3}{2})^2-(\frac{3}{2})^2+2} \,dx=\int_{0}^{k} \cfrac{1}{(x+\frac{3}{2})^2-\frac{9}{4}+2} \,dx$

$=\int_{0}^{k} \cfrac{1}{(x+\frac{3}{2})^2-\frac{1}{4}}\, dx=\int_{0}^{k} \cfrac{1}{(x+\frac{3}{2})^2-(\frac{1}{2})^2}\, dx$

$=\cfrac{1}{2\times\frac{1}{2}}\ln\left|\cfrac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2}}\right|\Big|_0^k=\boxed{\ln\left|\cfrac{k+1}{k+2}\right|-\ln\cfrac{1}{2}}$

Acum calculam acea suma:

$\sum_{k=1}^{n}a_k=\sum_{k=1}^{n} \left(\ln\left|\cfrac{k+1}{k+2}\right|-\ln\cfrac{1}{2}\right)=\ln\cfrac{2}{3}-\ln\cfrac{1}{2}+\ln\cfrac{3}{4}-\ln\cfrac{1}{2}+\ln\cfrac{4}{5}-\ln\cfrac{1}{2}+...+\ln\cfrac{n+1}{n+2}-\ln\cfrac{1}{2}$

Suma este de la 1 la n, deci se scade $\ln\cfrac{1}{2}$ de n ori, deci putem scrie:

$=\ln\cfrac{2}{3}+\ln\cfrac{3}{4}+\ln\cfrac{4}{5}+...+\ln\cfrac{n+1}{n+2}-n\ln\cfrac{1}{2}$

$=\ln\left(\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{4}{5}\times ...\times \cfrac{n+1}{n+2}\right)-n\ln\cfrac{1}{2}=\boxed{\ln\cfrac{2}{n+2}-n\ln\cfrac{1}{2}}$

Acum putem trece la calcularea limitei:

$\lim_{n\to\infty} \cfrac{n\ln\cfrac{1}{2}+\ln\cfrac{2}{n+2}-n\ln\cfrac{1}{2}}{\ln n}=\lim_{n\to\infty} \cfrac{\ln\cfrac{2}{n+2}}{\ln n}$

Se observa ca este vorba despre cazul de nedeterminare $-\frac{\infty}{\infty}$ deci putem folosi L'Hopital, adica derivam numaratorul si numitorul:

$=\lim_{n\to\infty} \cfrac{\cfrac{1}{\cfrac{2}{n+2}}\left(\cfrac{2}{n+2}\right)'}{\cfrac{1}{n}}=\lim_{n\to\infty}\cfrac{\cfrac{n+2}{2}\times\cfrac{-2}{(n+2)^2}}{\cfrac{1}{n}}=\lim_{n\to\infty}\cfrac{-\cfrac{1}{n+2}}{\cfrac{1}{n}}$

$=\boxed{\lim_{n\to\infty}\left(-\cfrac{n}{n+2}\right)=-1}$

Deci raspunsul corect este c).