Animated Solution for Mathematics - Definite Integration: ${\lim }_{n\to \infty }\left(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+...+\frac{1}{2n}\right)$ is equal to :-

From Discrete Sums to Continuous Beauty

Imagine standing before a mountain of numbers. You are asked to sum them up, but there is a catch: the number of terms is growing to infinity.

This is the essence of the problem:

At first glance, it looks like a chaotic mess. The secret, my friend, lies not in brute force, but in the elegant transformation of discrete sums into continuous integrals.

Decoding the DNA of the Series

Every complex series has a pattern, a hidden DNA. Let us look at the terms: $\frac{1}{1+n},\frac{1}{2+n},\frac{1}{3+n},\dots ,\frac{1}{n+n}$.

Notice the denominator? It is always $n$ plus some integer $r$. The numerator is always $1$.

So, the general term is $T_r=\frac{1}{n+r}$. As we count from the first term to the last, $r$ starts at $1$ and ends at $n$. We have successfully condensed this expression into a compact summation:

The Riemann Transformation

Now, we need to invoke the magic of the Riemann Sum. The goal is to force our expression into the form $\frac{1}{n}\sum f(\frac{r}{n})$.

Because this specific form is the definition of a definite integral, let us manipulate our summation. If we factor out an $n$ from the denominator, we get:

Now, our expression becomes:

We have the $\frac{1}{n}$ outside, and a function of $\frac{r}{n}$ inside. We are ready for the transformation.

The Calculus Bridge

This is where the math truly sings. We replace the discrete sum with a continuous integral.

The $\frac{1}{n}$ becomes our differential $dx$, and the ratio $\frac{r}{n}$ becomes our variable $x$. Our function is $f(x)=\frac{1}{1+x}$.

The lower limit is the value of $\frac{r}{n}$ as $n\to \infty$ for the first term ($r=1$), which is ${\lim }_{n\to \infty }\frac{1}{n}=0$. The upper limit is the value of $\frac{r}{n}$ for the last term ($r=n$), which is ${\lim }_{n\to \infty }\frac{n}{n}=1$.

We have transformed a terrifying infinite sum into the beautiful, simple integral:

The Final Elegance

We are almost there. The integral of $\frac{1}{1+x}$ is a fundamental result: $\ln (1+x)$.

We evaluate this from $0$ to $1$:

Subtracting the two, we arrive at our final answer: $\ln 2$.

It is a moment of pure clarity. What seemed like an impossible, infinite addition was simply the area under the curve $y=\frac{1}{1+x}$ from $0$ to $1$. Keep this tool in your arsenal, and no series will ever intimidate you again.