\(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{1^2 + 2^2 + 3^2 + .....n^2}{n^3}\)  किसके बराबर है?

संकल्पना:

\({1^2 + 2^2 + 3^2 + .....n^2} = \frac {n\ (n\ +\ 1)(2n\ +\ 1)}{6}\)

गणना:

दिया गया:

\(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{1^2 + 2^2 + 3^2 + .....n^2}{n^3}\)

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n\ (n\ +\ 1)(2n\ +\ 1)}{6\ n^3}\)

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n\ (n\ +\ 1)(2n\ +\ 1)}{6\ n^3}\)

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{(n\ +\ 1)(2n\ +\ 1)}{6\ n^2}\)

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n^2(1\ +\ \frac {1}{n})(2\ +\ \frac {1}{n})}{6\ n^2}\)

= \( {\left (1\ +\ \frac {1}{\infty}\right ) \left (2\ +\ \frac {1}{\infty }\right )}\ \frac 1 6\)

= \(\frac 1 3\)