lim (n→∞) (n/(n²+1²) + n/(n²+2²) + ... + 1/(5n)) is equal to : (1) π/4 (2) tan⁻¹(2) (3) tan⁻¹(3) (4)

Q: Why does the summation go up to $2n$? I thought it usually goes to $n$.

That is a brilliant observation! In many textbook problems, the sum ends at $n$, leading to an integral from $0$ to $1$. But here, the last term is $\frac{1}{5n}$. By rewriting it as $\frac{n}{n^2 + (2n)^2}$, we see the variable $r$ must reach $2n$ to match the pattern. Always trust the last term—it dictates the upper limit!

Q: Is it always valid to just replace $\frac{r}{n}$ with $x$ and $\frac{1}{n}$ with $dx$?

Yes, provided the limit is of the form $\lim_{n \to \infty} \sum \frac{1}{n} f(\frac{r}{n})$. This is the fundamental definition of the Riemann integral. It bridges the gap between discrete summation and continuous area. Think of it as 'zooming out' until the jagged rectangles of the sum smooth out into a perfect curve.

Q: What if the numerator wasn't $n$? Would the method still work?

If the numerator were a constant, say $1$, we would have to force an $n$ into the numerator and multiply by $\frac{1}{n}$ outside the sum to maintain the $\frac{1}{n}$ factor required for the $dx$ transformation. The Riemann sum method is incredibly robust; as long as you can isolate the $\frac{1}{n}$ and express the rest as a function of $\frac{r}{n}$, you can solve it!

Question

$\lim_{n 	o \infty} \left( \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \dots + \frac{1}{5n} ight)$ is equal to :

Options:

(A) $\frac{\pi}{4}$

(B) $	an^{-1}(2)$

(C) $	an^{-1}(3)$

(D) $\frac{\pi}{2}$

Sigma Percentile · Accepted Answer

### The Beauty of the Infinite Sum: From Chaos to Calculus

Imagine standing before a massive, intimidating wall of numbers. You see a series:

$$\lim_{n 	o \infty} \left( \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \dots + \frac{1}{5n} ight)$$

At first glance, it looks like a chaotic mess of fractions. But in the world of JEE Advanced, chaos is just order waiting to be discovered. Today, we are going to transform this discrete summation into a beautiful, continuous integral.

### Phase 1: Decoding the Hidden Pattern

The first step in any limit problem is to find the general term. We see terms like $\frac{n}{n^2+1^2}$ and $\frac{n}{n^2+2^2}$. It is clear that the general term is $T_r = \frac{n}{n^2+r^2}$.

But what about that last term, $\frac{1}{5n}$? It looks like an outsider. This is where your intuition must kick in. We need to force it to conform to our pattern.

If we multiply the numerator and denominator by $n$, we get $\frac{n}{5n^2}$. Now, split that $5n^2$ into $n^2 + 4n^2$. Suddenly, it becomes:

$$\frac{n}{n^2 + (2n)^2}$$

Do you see it? The pattern holds! The summation doesn't stop at $n$; it goes all the way to $r = 2n$. We have successfully decoded the boundary.

### Phase 2: The Riemann Transformation

Now, let us write this in the elegant language of sigma notation:

$$L = \lim_{n 	o \infty} \sum_{r=1}^{2n} \frac{n}{n^2+r^2}$$

To turn this into an integral, we need the magic factor of $\frac{1}{n}$. Let us factor out $n^2$ from the denominator:

$$L = \lim_{n 	o \infty} \sum_{r=1}^{2n} \frac{n}{n^2\left(1 + \left(\frac{r}{n}ight)^2ight)}$$

The $n$ in the numerator cancels with one $n$ in the denominator, leaving us with:

$$L = \lim_{n 	o \infty} \sum_{r=1}^{2n} \frac{1}{n} \cdot \frac{1}{1 + \left(\frac{r}{n}ight)^2}$$

This is the moment of truth. We have our $\frac{1}{n}$ (which will become $dx$) and our function of $\frac{r}{n}$ (which will become $f(x)$).

### Phase 3: The Integral Setup

We are now ready to bridge the gap between the discrete and the continuous. We define $f(x) = \frac{1}{1+x^2}$.

The lower limit is the limit of $\frac{r}{n}$ as $n 	o \infty$ for $r=1$, which is $0$. The upper limit is the limit of $\frac{r}{n}$ as $n 	o \infty$ for $r=2n$, which is $2$.

Thus, our infinite sum collapses into the definite integral:

$$L = \int_0^2 \frac{1}{1+x^2} dx$$

### Phase 4: The Final Calculation

We have arrived at the home stretch. The integral of $\frac{1}{1+x^2}$ is a standard result: $	an^{-1}(x)$.

Applying the Fundamental Theorem of Calculus, we evaluate this from $0$ to $2$:

$$L = [	an^{-1}(x)]_0^2 = 	an^{-1}(2) - 	an^{-1}(0)$$

Since $	an^{-1}(0) = 0$, our final answer is simply **$	an^{-1}(2)$**.

Look at what we have achieved. We took a complex, infinite summation and reduced it to a single, elegant value. This is the power of calculus—it allows us to see the underlying geometry of the universe.

Animated Solution for Mathematics - Definite Integration: $lim_{n \to \infty} (\frac{n}{n ^{2} + 1 ^{2}} + \frac{n}{n ^{2} + 2 ^{2}} + \dots + \frac{1}{5 n})$ is equal to :

Select Answer:

Visualized Solution

Analyze the Series Structure

Decoding the Last Term

Expressing in Sigma Notation

Preparing for Integration

The Transformation Rule

Finding Integral Limits

Setting up the Integral

Evaluation

Final Answer

The Sigma Insight: Definite Integral as a Limit of a Sum

The Beauty of the Infinite Sum

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​(n2+12n​+n2+22n​+⋯+5n1​) is equal to :