lim(n→∞) [n²/((n²+1)(n+1)) + n²/((n²+4)(n+2)) + n²/((n²+9)(n+3)) + ... + n²/((n²+n²)(n+n))] is equal

Q: Why did we divide by $n^3$ specifically? Why not $n^2$ or $n^4$?

Great question! In a Riemann sum, we need to isolate a $\frac{1}{n}$ term to represent the differential $dx$. Our general term is $\frac{n^2}{(n^2+r^2)(n+r)}$. If you look at the degrees, the denominator has a degree of $2+1=3$. To make the terms inside the denominator look like $(1 + (r/n)^2)$ and $(1 + r/n)$, we need to divide by $n^2$ and $n$ respectively. That is a total of $n^3$. It is all about matching the degree to create the dimensionless ratio $\frac{r}{n}$.

Q: I always get confused with the limits of integration. How do we know it's 0 to 1?

Think about the summation index $r$. It runs from $r=1$ to $r=n$. In the Riemann sum definition, our variable $x$ is defined as $\frac{r}{n}$. When $r=1$, $x = \frac{1}{n}$, and as $n \to \infty$, $x \to 0$. When $r=n$, $x = \frac{n}{n} = 1$. Thus, the summation covers the interval from $0$ to $1$.

Q: Partial fractions are so tedious. Is there a shortcut?

I understand the frustration! Partial fractions can feel like busy work, but they are the bridge between a complex rational function and standard integrals. There is no 'shortcut' that avoids the algebra, but you can use the 'cover-up' method to find $C$ quickly. For $A$ and $B$, comparing coefficients is usually the safest path to avoid errors. Embrace the process; it is a test of patience and precision, which is exactly what JEE Advanced demands.

Question

$\lim_{n 	o \infty} \left( \frac{n^2}{(n^2+1)(n+1)} + \frac{n^2}{(n^2+4)(n+2)} + \frac{n^2}{(n^2+9)(n+3)} + \dots + \frac{n^2}{(n^2+n^2)(n+n)} ight)$ is equal to

Options:

(A) $\frac{\pi}{8} + \frac{1}{4} \log_e 2$

(B) $\frac{\pi}{4} + \frac{1}{8} \log_e 2$

(C) $\frac{\pi}{4} - \frac{1}{8} \log_e 2$

(D) $\frac{\pi}{8} + \log_e \sqrt{2}$

Sigma Percentile · Accepted Answer

### The Beauty of the Infinite Sum: A Journey into Riemann

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of terms.

You see a series:

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

It is intimidating, isn't it? It feels like a mountain of algebra. But in the world of JEE Advanced, we don't climb mountains by brute force; we climb them by finding the hidden path. This problem is a classic example of a **Riemann Sum**, a beautiful bridge between discrete summation and continuous integration.

### Phase 1: Decoding the Pattern

Before we touch any calculus, we must understand the DNA of this series. Look at the general term, $T_r$.

The numerator is stubbornly constant at $n^2$. The denominator is a dance of two factors: $(n^2 + r^2)$ and $(n+r)$, where $r$ ranges from $1$ to $n$.

So, our general term is:

$$T_r = \frac{n^2}{(n^2+r^2)(n+r)}$$

This is our building block. Without identifying this, we are lost. With it, we have the key to the kingdom.

### Phase 2: The Surgical Strike

Now, we need to transform this discrete sum into an integral. The definition of a Riemann sum is:

$$\lim_{n 	o \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}ight) = \int_0^1 f(x) \, dx$$

To get there, we need a $\frac{1}{n}$ factor. We perform a surgical strike on our general term by dividing both the numerator and the denominator by $n^3$.

By distributing $n^2$ to the first bracket and $n$ to the second, we obtain:

$$T_r = \frac{1/n}{(1 + (r/n)^2)(1 + r/n)}$$

Suddenly, the chaos vanishes. We have a function of $\frac{r}{n}$ multiplied by $\frac{1}{n}$. This is the moment of clarity.

### Phase 3: The Integral Transformation

As $n$ approaches infinity, the sum of these rectangles becomes the area under the curve $f(x) = \frac{1}{(1+x^2)(1+x)}$. The summation symbol $\sum$ becomes the integral sign $\int$, the $\frac{r}{n}$ becomes $x$, and the $\frac{1}{n}$ becomes $dx$.

Our limits are $0$ to $1$. We are now solving:

$$\int_0^1 \frac{1}{(1+x^2)(1+x)} \, dx$$

We have turned a terrifying infinite series into a standard definite integral. This is the elegance of calculus.

### Phase 4: The Algebraic Heavy Lifting

Now, we face the integral. We use **partial fraction decomposition**:

$$\frac{1}{(1+x^2)(1+x)} = \frac{Ax+B}{1+x^2} + \frac{C}{1+x}$$

By multiplying through by the denominator and solving for the constants, we find $C = \frac{1}{2}$, $A = -\frac{1}{2}$, and $B = \frac{1}{2}$.

This breaks our integral into three manageable parts:

$$\frac{1}{2} \int_0^1 \left( \frac{1}{1+x^2} - \frac{x}{1+x^2} + \frac{1}{1+x} ight) \, dx$$

### Phase 5: The Final Victory

We integrate term by term. The first is $	an^{-1}(x)$, the second is $-\frac{1}{2} \ln(1+x^2)$, and the third is $\ln(1+x)$.

Evaluating from $0$ to $1$:

$$\frac{1}{2} \left[ \left( \frac{\pi}{4} - \frac{1}{2} \ln(2) + \ln(2) ight) - (0 - 0 + 0) ight]$$

Simplifying this, we arrive at the final result:

$$\mathbf{\frac{\pi}{8} + \frac{1}{4} \ln(2)}$$

You did it. You took a complex, intimidating series and, through the power of Riemann sums and partial fractions, reduced it to a clean, elegant result. This is not just math; this is mastery.

Animated Solution for Mathematics - Definite Integration: $lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})$ is equal to

Select Answer:

Visualized Solution

Analyze the Series Pattern

Identify the General Term $T_{r}$

Transform into Riemann Sum Form

Simplify the General Term

Convert Sum to Definite Integral

The Definite Integral

Partial Fraction Decomposition

Solve for Constants

Rewrite the Integral

Split into Standard Integrals

Integrate Term by Term

Apply the Upper Limit ( $x = 1$ )

Final Calculation

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

Phase 5

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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+1)(n+1)n2​+(n2+4)(n+2)n2​+(n2+9)(n+3)n2​+⋯+(n2+n2)(n+n)n2​) is equal to