lim n→∞ (1 + 1/n² + 1/n² + ... + 1/n)ⁿ is equal to : (1) e^(1/2) (2) e^(1/e) (3) 1 (4) 0

Q: Why is the limit $1^\infty$ indeterminate? Shouldn't $1$ to any power be $1$?

That is a classic trap! In calculus, $1^\infty$ is not a fixed value of $1$. It represents a base that is *approaching* $1$ while the exponent is *approaching* infinity. Because the base is slightly larger than $1$ (e.g., $1.0001$), raising it to a massive power can result in any value, including $1$, $e^k$, or even infinity. We must use the limit identity $e^{\lim f(n)g(n)}$ to resolve this competition between the base and the exponent.

Q: How can we just replace $H_n$ with $\ln n$? Is that mathematically rigorous?

Great question! While $H_n \approx \ln n + \gamma$ (where $\gamma$ is the Euler-Mascheroni constant) is an approximation, it is perfectly rigorous for evaluating the limit. Since we are looking at the behavior as $n \to \infty$, the constant $\gamma$ becomes negligible compared to $\ln n$. The ratio $\frac{\ln n + \gamma}{n}$ behaves exactly like $\frac{\ln n}{n}$ in the limit, which is $0$.

Q: Can I use L'Hopital's rule on the harmonic sum?

L'Hopital's rule applies to functions of a continuous variable $x$. The harmonic sum $H_n$ is a discrete sum. However, you can approximate $H_n$ using the integral $\int_1^n \frac{1}{x} dx = \ln n$. Using this continuous approximation, you can indeed apply L'Hopital's rule to $\frac{\ln x}{x}$, which gives $\lim_{x \to \infty} \frac{1/x}{1} = 0$. It confirms our result beautifully!

Question

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

Options:

(A) $e^{1/2}$

(B) $e^{1/e}$

(C) $1$

(D) $0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with the limit: 
$$\lim_{n 	o \infty} \left( 1 + \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n^2} ight)^n$$

The first thing that should catch your eye is the structure. We have a base that looks like $1 + 	ext{something}$ and an exponent that is $n$.

As $n$ grows, the term $H_n = 1 + \frac{1}{2} + \dots + \frac{1}{n}$ divided by $n^2$ shrinks. Since $H_n$ grows logarithmically and $n^2$ grows quadratically, the fraction $\frac{H_n}{n^2}$ vanishes to $0$.

Thus, our base approaches $1$ and our exponent approaches $\infty$. We have successfully identified a **$1^\infty$ indeterminate form**.

### The Transformation

Whenever you see a $1^\infty$ form, your mind should immediately jump to the standard limit property: 
$$\lim_{x 	o a} (1 + f(x))^{g(x)} = e^{\lim_{x 	o a} f(x)g(x)}$$

This is our golden key. Here, our $f(n)$ is $\frac{H_n}{n^2}$ and our $g(n)$ is $n$.

By applying this, the limit transforms into:
$$e^{\lim_{n 	o \infty} n \cdot \left( \frac{H_n}{n^2} ight)}$$

We have moved the complexity from the exponent of the base into a simple product in the exponent of $e$.

### The Battle of Growth Rates

Now, let us focus on the exponent: 
$$E = \lim_{n 	o \infty} n \cdot \frac{H_n}{n^2}$$

A simple algebraic cancellation of $n$ leaves us with:
$$E = \lim_{n 	o \infty} \frac{H_n}{n}$$

This is where the growth rate becomes the star of the show. We know that the harmonic sum $H_n$ behaves like $\ln n$ for large $n$. So, we are effectively evaluating $\lim_{n 	o \infty} \frac{\ln n}{n}$.

Imagine a race. In the numerator, we have $\ln n$, which climbs steadily but sluggishly. In the denominator, we have $n$, which is a linear function, shooting upwards with constant velocity.

As $n$ approaches infinity, the linear growth of the denominator completely dominates the logarithmic growth of the numerator. The gap between them widens until the ratio is crushed to zero.

Mathematically:
$$\lim_{n 	o \infty} \frac{\ln n}{n} = 0$$

### Final Calculation

We have found that our exponent $E$ is $0$. Bringing this back to our base $e$, we get:
$$L = e^0$$

As every student of mathematics knows, any non-zero number raised to the power of zero is $1$.

The complexity of the harmonic sum, the quadratic denominator, and the infinite exponent all collapse into the simple, elegant result:
**Result = 1**

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{n \to \infty} (1 + \frac{1}{n ^{2}} + \frac{1}{n ^{2}} + ....... + \frac{1}{n})^{n}$ is equal to :

Select Answer:

Visualized Solution

Analyze the Limit Structure

Identify the Indeterminate Form

Apply the $1^{\infty}$ Transformation

Simplify the Exponent

Bound the Harmonic Sum

Evaluate the Exponent Limit

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Transformation

The Battle of Growth Rates

Final Calculation

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