Question

${lim}_{n\to \infty }\frac{1+1+4+8+9+27+16+64+...}{(1+2+3+4+5+...{)}^2}=$............

• A. $1$
• B. $0$
• C. $infinity$
• D. $noneofthese$

Answer 1

The correct option is A $1$

${lim}_{n\to \infty }\frac{(1+4+9+16+...)}{(1+2+3+4+5{)}^2}+\frac{(1+8+27+...64+...)}{(1+2+3+...{)}^2}$

$\frac{n(n+1)(2n+1)}{6\frac{\left(n{)}^2\right(2n+1{)}^2}{4}}+\frac{(\frac{n(n+1)}{2}{)}^2}{(\frac{n\left(n\right(1\left)\right)}{2}{)}^2}$

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

$=1$ Ans