Animated Solution for Mathematics - Definite Integration: Among $(S1):{\lim }_{n\to \infty }\frac{1}{n^2}(2+4+6+\cdots +2n)=1$ and $(S2):{\lim }_{n\to \infty }\frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\cdots +n^{15})=\frac{1}{16}$

The Arithmetic of Infinity

Let us start with $(S1)$:

The numerator is a classic arithmetic progression. If we factor out a $2$, we get $2(1+2+3+\cdots +n)$.

We know the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$. Substituting this, the numerator becomes:

Now, our limit is simply:

Dividing by $n^2$, we get ${\lim }_{n\to \infty }(1+\frac{1}{n})$. As $n$ marches toward infinity, $\frac{1}{n}$ vanishes into zero, leaving us with exactly $1$. Therefore, Statement $(S1)$ is true.

Unlocking (S2)

Now, let us tackle $(S2)$:

This looks intimidating, but it is a classic setup for the Riemann sum. We can rewrite this expression as:

We split $n^{16}$ into $n\times n^{15}$ because we need that $\frac{1}{n}$ factor to represent the width of our rectangles in the integral definition.

Imagine the graph of $y=x^{15}$. We are dividing the interval $[0,1]$ into $n$ tiny rectangles, each with a width of $\frac{1}{n}$.

The height of the $r$-th rectangle is $(\frac{r}{n}{)}^{15}$. As $n$ approaches infinity, the sum of the areas of these infinitely thin rectangles becomes the area under the curve $y=x^{15}$ from $x=0$ to $x=1$.

This is the magic of the definite integral:

Evaluating this integral is a joy. Using the power rule, $\int x^{n}dx=\frac{x^{n+1}}{n+1}$, we find:

Plugging in the limits, we get:

It matches perfectly! Both statements are true. Remember, in JEE, the most complex-looking problems often hide the most elegant, simple truths.