CUET Preparation Today
$\underset{n→∞}{\lim}\sum\sum\limits_{1≤i<j≤n}\frac{(i+1)(j+1)}{n^4}$ is |
$\frac{1}{4}$ $\frac{1}{6}$ $\frac{1}{8}$ none of these |
$\frac{1}{8}$ |
$\underset{n→∞}{\lim}\frac{\left(\sum\limits_{i=1}^{n}(i+1)\right)^2=\sum\limits_{i=1}^{n}(i+1)^2}{2n^4}$ $=\underset{n→∞}{\lim}\frac{\left(\frac{(n+2)(n+1)}{2}-1\right)^2-\left(\frac{(n+1)(n+2)(2n+3)}{6}\right)}{2n^4}=\frac{1}{8}$ |
