Problem 182 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find \(\lim _{n \rightarrow \infty} n^{-n^{2}}\left((n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^{2}}\right) \cdots\left(n+\frac{1}{2^{n-1}}\right)\right)^{n}\)

Step-by-Step Solution

Answer

The given expression \(\lim _{n \rightarrow \infty} n^{-n^{2}}\left((n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^{2}}\right) \cdots\left(n+\frac{1}{2^{n-1}}\right)\right)^{n}\) does not have a limit as \(n\) approaches infinity.

Step 1: Simplify the exponent

We can rewrite the expression as: \[ \lim_{n \rightarrow \infty} \exp\left(-n^2 \log n + n \log\left(\left(n + 1\right)\left(n + \frac{1}{2}\right)\left(n + \frac{1}{2^2}\right) \cdots \left(n + \frac{1}{2^{n-1}}\right)\right)\right) \] We used the property: \(\log a^b = b \log a\). So, we simplified \(n^{-n^2}\left(\cdots\right)^n\) as \(e^{-n^2 \log n + n\log(\cdots)}\).

Step 2: Simplify the logarithmic expression

We can further simplify the logarithmic expression by using the property \(\log(a \cdot b) = \log a + \log b\). The expression becomes: \[ \lim_{n \rightarrow \infty} \exp\left(-n^2 \log n + n\sum_{k=0}^{n-1} \log\left(n + \frac{1}{2^k}\right)\right) \] Now, let's focus on simplifying the sum of logarithms inside the limit:

Step 3: Apply Squeeze theorem

To apply the Squeeze theorem on the sum, consider the following inequalities for \(0\leq k\leq n-1\): \[ \log\left(n\right) \leq \log\left(n + \frac{1}{2^k}\right) \leq \log\left(n+1\right) \] Multiplying by \(n\) on both sides: \[ n\log\left(n\right) \leq n\sum_{k=0}^{n-1} \log\left(n + \frac{1}{2^k}\right) \leq n\log\left(n+1\right) \]

Step 4: Compute the limits of the bounds

Compute the limits as n approaches infinity for both bounds: \[ \lim_{n \rightarrow \infty} n\log\left(n\right)=\infty \] \[ \lim_{n \rightarrow \infty} n\log\left(n+1\right)=\infty \] Since the sum is between two quantities that tend to infinity, then the sum also tends to infinity: \[ \lim_{n \rightarrow \infty} n\sum_{k=0}^{n-1} \log\left(n + \frac{1}{2^k}\right) = \infty \]

Step 5: Find the final limit

Now that the sum tends to infinity, we can find the limit of the given expression: \[ \lim_{n \rightarrow \infty} \exp\left(-n^2 \log n + n\sum_{k=0}^{n-1} \log\left(n + \frac{1}{2^k}\right)\right) = \exp\left(\infty - \infty\right) \] Since the exponent is in indeterminate form \(\infty - \infty\), we cannot evaluate this limit directly. The given expression does not have a limit as \(n \to \infty\).

Key Concepts

Squeeze Theorem

The Squeeze Theorem is an essential tool in calculus that helps find limits. It's especially useful when direct evaluation is tough.
Here's how it works: if you have three functions, say \( f(n) \), \( g(n) \), and \( h(n) \), and they satisfy \( f(n) \leq g(n) \leq h(n) \) for all \( n \) in some interval, and if the limits of \( f(n) \) and \( h(n) \) as \( n \to \infty \) are the same, then the limit of \( g(n) \) as \( n \to \infty \) will also be the same.
This theorem is like a sandwich. If two slices of bread (\( f(n) \) and \( h(n) \)) are going to the same place, the filling (\( g(n) \)) has to go there too.
In solving complex limits, like in our exercise, the Squeeze Theorem can confirm tail behavior when direct analysis seems tough.

Logarithmic Properties

Logarithmic properties help in simplifying complex expressions in calculus. They are like rules that guide us in doing algebra with logs.
Here are some key properties:

• \( \log(ab) = \log a + \log b \)
• \( \log(a^b) = b \log a \)
• \( \log\left(\frac{a}{b}\right) = \log a - \log b \)

Using these properties, you can turn products into sums, powers into products, and quotients into differences.
This simplification is crucial, as seen in our exercise when we convert products inside logarithmic functions to sums, making them easier to manage in limits.

Indeterminate Forms

In calculus, an indeterminate form arises when trying to find a limit that doesn't lead to any particular number directly, due to conflicting operations.
Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and \( \infty - \infty \).
When you see these forms, further steps or transformations are required to make sense of the limit.
For example, the original exercise presents an \( \infty - \infty \) form when calculating the exponent of an expression, which prevents direct evaluation of the limit.

Infinity in Calculus

Infinity is a concept used in calculus to explore limits and behaviors of functions as they grow large or small.
We often deal with 'approaches' rather than actual numbers, like when \( n \to \infty \), indicating something is growing without bound.
This is key in limit problems, where expressions often become complex as a variable grows towards infinity.
In our exercise, examining how various terms grow as \( n \) increases helps understand the behavior of the limit.
Despite being abstract, infinity serves as a useful way to test and explore calculations beyond finite confines.