Animated Solution for Mathematics - Definite Integration: ${\lim }_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n}{e}^{r/n}$ is

Welcome, fellow traveler on the path of calculus! Today, we are going to demystify an expression that often makes students pause:

At first glance, it looks like a daunting, infinite mountain of terms. But I want you to see it not as a burden, but as a beautiful, geometric landscape waiting to be measured.

Imagine you are standing before the curve $y=e^x$. You want to find the area under this curve from $x=0$ to $x=1$. How do you do it? You draw rectangles!

If you draw $n$ rectangles, each with a width of $\frac{1}{n}$, and you sum their areas, you get an approximation. As $n$ grows larger and larger, these rectangles become thinner and thinner, eventually filling the space under the curve perfectly.

This is the essence of the Riemann Sum. Our expression is simply the mathematical shorthand for this process as $n$ approaches infinity.

To solve this, we must translate the language of summation into the language of integration. Think of it as a dictionary for calculus:

1. The width of our tiny rectangles, $\frac{1}{n}$, transforms into the differential element $dx$. 2. The position variable, $\frac{r}{n}$, becomes our continuous variable $x$. 3. The limit of the summation, ${\lim }_{n\to \infty }\sum$, transforms into the integral sign $\int$.

By applying these rules, our intimidating sum $\sum \frac{1}{n}{e}^{r/n}$ becomes the much friendlier integral of $e^{x}dx$.

Now, we need to know where our area starts and ends. We look at the bounds of our summation index $r$.

The lower bound is $r=1$. As $n\to \infty$, the ratio $\frac{r}{n}$ becomes $\frac{1}{n}$, which approaches $0$.

The upper bound is $r=n$. As $n\to \infty$, the ratio $\frac{r}{n}$ becomes $\frac{n}{n}$, which is $1$.

Thus, our integral is defined from $0$ to $1$. We have successfully collapsed the infinite sum into the definite integral:

We are now at the finish line. The function $e^x$ is one of the most elegant functions in mathematics because its antiderivative is simply itself.

We evaluate the integral:

Applying the Fundamental Theorem of Calculus, we substitute the upper limit and subtract the value at the lower limit:

Since $e^0=1$, our result is $e-1$.

There you have it! We started with an infinite sum and ended with a precise, elegant value. You have just mastered the Riemann Sum. Keep this intuition close, and you will find that even the most complex limits become clear when you visualize the geometry behind them.