Question

${lim}_{n\to \infty }\frac{2^3-1^3}{2^3+1^3}\cdot \frac{3^3-1^3}{3^3+1^3}\dots \frac{n^3-1}{n^3+1}$ equals

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$\frac{2}{3}$

We have,

${lim}_{n\to \infty }\frac{2^3-1^3}{2^3+1^3}$$.\frac{3^3-1^3}{3^3+1^3}$$.....\frac{n^3-1}{n^3+1}$

Here general term $\frac{n^3-1}{n^3+1}=\frac{(n-1)\left[\right(n+1{)}^2-(n+1)+1]}{(n+1)(n^2-n+1)}$

Putting values at $n=1,2,3,...n$

$={lim}_{n\to \infty }=\frac{2}{3}\frac{(n^2+n+1)}{n(n+1)}=\frac{2}{3}$

Hence, this is the answer.