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Sa se arate ca

\( {\lim}\limits_{n\to\infty}[a\ln(3+n)+b\ln(2+n)+c\ln(1+n)]=0 <=> a+b+c=0 \)

Admitere, Informatica, Univ. "A.I.Cuza" Iasi, 1997

\( \lim_{n\rightarrow\infty}[a\ln(3+n)+b\ln(n+2)+c\ln(n+1)]=\lim_{n\rightarrow\infty}[a\ln(1+\frac{3}{n})+b\ln(1+\frac{2}{n})+c\ln(1+\frac{1}{n})+a\ln{n}+b\ln{n}+c\ln{n}]=\lim_{n\rightarrow\infty}[a\frac{3}{n}+b\frac{2}{n}+c\frac{1}{n}+(a+b+c)\ln{n}]=\lim_{n\rightarrow\infty}[(a+b+c)\ln{n}] \)
\( \lim_{n\rightarrow\infty}[a\ln(3+n)+b\ln(2+n)+c\ln(1+n)]=0 <=>\lim_{n\rightarrow\infty}[(a+b+c)\ln{n}]=0<=> {a+b+c}=0 \)

Totul e corect in afara de faptul ca te-ai complicat scriind ca \( \lim_{x\to 0}ln(1+x)=\lim_{x\to 0} x \) ceea ce este evident caci sunt 0 amandoua. Puteai pur si simplu sa inlocuiesti cu 0 acolo deoarece e o suma finita.

Notam \( a_n= a\ln(3+n)+b\ln(2+n)+c\ln(1+n) \) . Se arata usor ca

\( a_n=(a+b+c)\cdot\ln n+a\ln\left(1+\frac 3n\right)+b\ln\left(1+\frac 2n\right)+c\ln\left(1+\frac 1n\right) \) .

Se observa ca \( \lim_{n\to\infty}\ \left[a\ln\left(1+\frac 3n\right)+b\ln\left(1+\frac 2n\right)+c\ln\left(1+\frac 1n\right)\right]=0 \)

si \( \lim_{n\to\infty}\ a_n=\lim_{n\to\infty}\ (a+b+c)\cdot\ln n=\left\{\begin{array}{ccc}
-\infty & \mathrm{daca} & a+b+c<0\\\\\\\\
0 & \mathrm{daca} & a+b+c=0\\\\\\\\
\infty & \mathrm{daca} & a+b+c>0\end{array} \) .