If U_n = (1 + 1/n²)(1 + 2²/n²)² ... (1 + n²/n²)^n, then lim_{n→∞} (U_n)^(-4/n²) is equal to : (1) e²

Q: Why do we take the natural logarithm at the start? Can't we just evaluate the limit directly?

Great question! The expression is a product of $n$ terms, and as $n$ approaches infinity, the number of terms also approaches infinity. Evaluating the limit of a product directly is extremely difficult because the terms themselves are changing with $n$. Taking the natural logarithm, $\ln$, transforms the product into a summation, which is the standard form for a Riemann sum. It turns a 'product of doom' into a 'sum of manageable terms'.

Q: Why does the exponent $r$ inside the product become a multiplier in the sum?

This is a direct application of the logarithmic property $\ln(a^b) = b \ln a$. When we take the log of the product $\prod (1 + \frac{r^2}{n^2})^r$, the log of the product becomes the sum of the logs: $\sum \ln((1 + \frac{r^2}{n^2})^r)$. Then, using the power rule, that exponent $r$ comes down in front of the log, giving us $r \ln(1 + \frac{r^2}{n^2})$. It is a beautiful simplification!

Q: How do I know when to use the Riemann sum method?

Look for the signature of a Riemann sum: a limit as $n \to \infty$ of a summation $\sum_{r=1}^n$ where the terms inside can be expressed as a function of $\frac{r}{n}$, multiplied by a $\frac{1}{n}$ factor. If you see this structure, you are almost certainly dealing with a Riemann sum problem. It is the bridge between discrete summation and continuous integration.

Q: Why do the limits of integration change from $0$ to $1$?

When we convert the sum to an integral, we replace $\frac{r}{n}$ with $x$. The lower limit of the sum is $r=1$, so $\frac{r}{n} = \frac{1}{n}$, which approaches $0$ as $n \to \infty$. The upper limit is $r=n$, so $\frac{r}{n} = \frac{n}{n} = 1$. Thus, the range of $x$ is from $0$ to $1$. When we perform the substitution $t = 1 + x^2$, we must update the limits again based on the new variable $t$.

Question

If $U_n = \left(1 + \frac{1}{n^2}ight) \left(1 + \frac{2^2}{n^2}ight)^2 \dots \left(1 + \frac{n^2}{n^2}ight)^n$, then $\lim_{n 	o \infty} (U_n)^{-4/n^2}$ is equal to :

Options:

(A) $\frac{e^2}{16}$

(B) $\frac{4}{e}$

(C) $\frac{16}{e^2}$

(D) $\frac{4}{e^2}$

Sigma Percentile · Accepted Answer

### The Product of Doom: A Journey into Limits

Imagine you are standing at the edge of a mathematical abyss. You are presented with a sequence $U_n$ that looks like a monster: a product of terms, each with its own exponent, all growing as $n$ marches toward infinity.

It is easy to feel overwhelmed, but in the world of JEE Advanced, every monster has a weakness. Our goal is to find $\lim_{n 	o \infty} (U_n)^{-4/n^2}$. Let us break this down together.

### Phase 1: The Logarithmic Bridge

We start with the expression:

$$U_n = \prod_{r=1}^n \left(1 + \frac{r^2}{n^2}ight)^r$$

Dealing with a product of $n$ terms where $n$ is going to infinity is a nightmare. But remember, logarithms are the great simplifiers; they turn products into sums.

Let $L = \lim_{n 	o \infty} (U_n)^{-4/n^2}$. Taking the natural logarithm of both sides, we get $\ln L = \lim_{n 	o \infty} \ln((U_n)^{-4/n^2})$.

Using the power rule for logarithms, the exponent $-\frac{4}{n^2}$ jumps to the front, and the log of the product becomes the sum of the logs:

$$\ln L = \lim_{n 	o \infty} -\frac{4}{n^2} \sum_{r=1}^n r \ln\left(1 + \frac{r^2}{n^2}ight)$$

Suddenly, the monster is much smaller.

### Phase 2: The Riemann Sum

Now, look at the structure of our sum. We have a $\frac{1}{n^2}$ outside and an $r$ inside. This is the classic setup for a Riemann sum.

We need to create terms of $\frac{r}{n}$ and a $\frac{1}{n}$ outside. Let us rewrite the expression:

$$\ln L = -4 \lim_{n 	o \infty} \sum_{r=1}^n \left(\frac{r}{n}ight) \ln\left(1 + \left(\frac{r}{n}ight)^2ight) \frac{1}{n}$$

This is the moment of truth. As $n 	o \infty$, this sum transforms into the definite integral:

$$\int_0^1 x \ln(1 + x^2) \, dx$$

Here, the $\frac{r}{n}$ becomes $x$, and the $\frac{1}{n}$ becomes $dx$. We have successfully bridged the gap from discrete summation to continuous calculus.

### Phase 3: The Final Integration

We are left with the integral $\ln L = -4 \int_0^1 x \ln(1 + x^2) \, dx$. This looks intimidating, but a simple substitution will save us.

Let $t = 1 + x^2$. Then $dt = 2x \, dx$, or $x \, dx = \frac{dt}{2}$. When $x=0$, $t=1$. When $x=1$, $t=2$.

The integral becomes:

$$\ln L = -4 \int_1^2 \ln(t) \frac{dt}{2} = -2 \int_1^2 \ln t \, dt$$

The integral of $\ln t$ is a standard result: $t \ln t - t$. Evaluating this from $1$ to $2$:

$$\ln L = -2 \left[ (2 \ln 2 - 2) - (1 \ln 1 - 1) ight]$$

Since $\ln 1 = 0$, this simplifies to:

$$\ln L = -2 [2 \ln 2 - 2 + 1] = -2 [2 \ln 2 - 1] = 2 - 4 \ln 2$$

### The Grand Finale

We have $\ln L = 2 - 4 \ln 2$. We can write $2$ as $\ln(e^2)$ and $4 \ln 2$ as $\ln(2^4) = \ln(16)$.

So, $\ln L = \ln(e^2) - \ln(16) = \ln\left(\frac{e^2}{16}ight)$. Taking the antilog, we find:

$$L = \frac{e^2}{16}$$

We have conquered the monster! The final answer is **$\frac{e^2}{16}$**. Remember, no matter how complex a limit looks, there is always a path to simplify it.

Animated Solution for Mathematics - Definite Integration: If $U_{n} = (1 + \frac{1}{n ^{2}}) (1 + \frac{2 ^{2}}{n ^{2}})^{2} \dots (1 + \frac{n ^{2}}{n ^{2}})^{n}$ , then $lim_{n \to \infty} (U_{n})^{- 4/ n^{2}}$ is equal to :

Select Answer:

Visualized Solution

Define the Product $U_{n}$

Apply Logarithms to the Limit

Transform Product into Summation

Simplify the Summation Expression

Convert to Riemann Sum Form

Transform Sum to Definite Integral

Substitution for Integration

Change Limits and Simplify

Integrate $ln t$

Evaluate the Definite Integral

Final Logarithmic Simplification

Final Answer

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

The Product of Doom

Phase 1

Phase 2

Phase 3

The Grand Finale

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