34. Determine The Limit Of The Sequence $u_n = \sum_{k=0}^{n-1} \frac{1}{k^2 + N^2}$ As N Approaches +∞. 35. Determine The Limit Of The Sequence $u_n = \prod_{k=1}^{n} (1 + \frac{k^2}{n^2})^{\frac{1}{n}}$ As N Approaches +∞. 36. Discussion Category: Mathematics.

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In this section, we embark on a mathematical journey to determine the limit of the sequence $u_n$ as $n$ approaches positive infinity, where $u_n$ is defined by the summation $u_n = \sum_{k=0}^{n-1} \frac{1}{k^2 + n^2}$. This problem delves into the realm of calculus and sequences, requiring a keen understanding of summation techniques and limit evaluation. To unravel the mystery of this limit, we'll employ a strategy that involves transforming the summation into a form that lends itself to analysis using integral calculus. This approach hinges on the concept of Riemann sums, which provide a bridge between discrete summations and continuous integrals.

Our primary objective is to express the given summation as a Riemann sum, which will allow us to approximate it using a definite integral. By carefully manipulating the expression, we aim to identify a suitable function and integration interval that accurately represent the summation's behavior as $n$ grows infinitely large. This transformation will enable us to leverage the power of integral calculus to evaluate the limit with precision. Let's begin by examining the structure of the summation and identifying potential avenues for conversion into a Riemann sum.

Delving into the structure of the summation, we observe that each term has the form $\frac{1}{k^2 + n^2}$. A key step in transforming this into a Riemann sum involves factoring out $n^2$ from the denominator. This manipulation yields $\frac{1}{n^2( (k/n)^2 + 1)}$. Now, we can rewrite the summation as follows:

$u_n = \sum_{k=0}^{n-1} \frac{1}{k^2 + n^2} = \sum_{k=0}^{n-1} \frac{1}{n^2( (k/n)^2 + 1)} = \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{n( (k/n)^2 + 1)} = \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{(k/n)^2 + 1}$.

This manipulation is crucial because it reveals the presence of the term $\frac{k}{n}$, which is a hallmark of Riemann sums. We now recognize that the summation closely resembles a Riemann sum for the function $f(x) = \frac{1}{x^2 + 1}$ over a suitable interval. To make this connection explicit, let's consider the interval $[0, 1]$ and divide it into $n$ equal subintervals of width $\frac{1}{n}$. The right endpoint of the $k$-th subinterval is given by $x_k = \frac{k}{n}$.

Thus, the Riemann sum for the function $f(x) = \frac{1}{x^2 + 1}$ over the interval $[0, 1]$ can be written as:

$\sum_{k=0}^{n-1} f(x_k) \Delta x = \sum_{k=0}^{n-1} \frac{1}{(k/n)^2 + 1} \cdot \frac{1}{n}$.

Comparing this with our expression for $u_n$, we see that:

$u_n = \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{(k/n)^2 + 1}$.

As $n$ approaches infinity, the Riemann sum converges to the definite integral of $f(x)$ over the interval $[0, 1]$. Therefore, we can write:

$\lim_{n \to +\infty} u_n = \lim_{n \to +\infty} \sum_{k=0}^{n-1} \frac{1}{k^2 + n^2} = \int_{0}^{1} \frac{1}{x^2 + 1} dx$.

Now, we need to evaluate the definite integral. Recall that the antiderivative of $\frac{1}{x^2 + 1}$ is the arctangent function, denoted as $\arctan(x)$. Therefore, we have:

$\int_{0}^{1} \frac{1}{x^2 + 1} dx = \arctan(x) \Big|_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

Thus, the limit of the sequence $u_n$ as $n$ approaches positive infinity is $\frac{\pi}{4}$. This result highlights the power of Riemann sums in connecting discrete summations with continuous integrals, allowing us to solve problems that might otherwise be intractable.

In summary, by recognizing the summation as a Riemann sum and leveraging the fundamental theorem of calculus, we successfully determined the limit of the sequence $u_n$ as $n$ approaches infinity. The process involved factoring, recognizing the Riemann sum structure, and evaluating a definite integral, showcasing the interplay between discrete and continuous mathematics.

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In this section, we embark on a captivating exploration to determine the limit of the sequence $u_n$ as $n$ tends towards positive infinity, where $u_n$ is defined by the product $u_n = \prod_{k=1}^{n} (1 + \frac{k^2}{n^2})^{\frac{1}{n}}$. This problem lies within the domain of calculus and sequences, requiring a sophisticated understanding of product manipulations, limits, and logarithms. To unravel the intricacies of this limit, we will employ a strategy centered around transforming the product into a summation using logarithms. This transformation will pave the way for us to leverage the power of Riemann sums and integral calculus to evaluate the limit with precision.

Our primary objective is to convert the given product into a summation by taking the natural logarithm of both sides of the equation. This transformation is motivated by the property of logarithms that converts products into sums. By applying the natural logarithm, we will effectively transform the product into a form that is more amenable to analysis using summation techniques. Once we have the summation, we will attempt to express it as a Riemann sum, which will allow us to approximate it using a definite integral. This approach mirrors the strategy employed in the previous problem, highlighting the versatility of Riemann sums in dealing with limits of discrete expressions.

Let's begin by taking the natural logarithm of $u_n$:

$\ln(u_n) = \ln \left( \prod_{k=1}^{n} \left(1 + \frac{k^2}{n^2} \right)^{\frac{1}{n}} \right)$.

Using the properties of logarithms, we can rewrite the expression as:

$\ln(u_n) = \sum_{k=1}^{n} \ln \left( \left(1 + \frac{k^2}{n^2} \right)^{\frac{1}{n}} \right) = \frac{1}{n} \sum_{k=1}^{n} \ln \left(1 + \frac{k^2}{n^2} \right)$.

Now, we have a summation that we can analyze. Our next goal is to express this summation as a Riemann sum. We observe the presence of the term $\frac{k}{n}$, which is a key indicator that a Riemann sum representation is possible. To make the connection explicit, let's consider the function $f(x) = \ln(1 + x^2)$. We can rewrite the summation as:

$\ln(u_n) = \frac{1}{n} \sum_{k=1}^{n} \ln \left(1 + \left(\frac{k}{n} \right)^2 \right)$.

This summation closely resembles a Riemann sum for the function $f(x) = \ln(1 + x^2)$ over the interval $[0, 1]$. To see this, let's divide the interval $[0, 1]$ into $n$ equal subintervals of width $\frac{1}{n}$. The right endpoint of the $k$-th subinterval is given by $x_k = \frac{k}{n}$. Thus, the Riemann sum for the function $f(x) = \ln(1 + x^2)$ over the interval $[0, 1]$ can be written as:

$\sum_{k=1}^{n} f(x_k) \Delta x = \sum_{k=1}^{n} \ln \left(1 + \left(\frac{k}{n} \right)^2 \right) \cdot \frac{1}{n}$.

Comparing this with our expression for $\ln(u_n)$, we see that:

$\ln(u_n) = \frac{1}{n} \sum_{k=1}^{n} \ln \left(1 + \left(\frac{k}{n} \right)^2 \right)$.

As $n$ approaches infinity, the Riemann sum converges to the definite integral of $f(x)$ over the interval $[0, 1]$. Therefore, we can write:

$\lim_{n \to +\infty} \ln(u_n) = \lim_{n \to +\infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(1 + \left(\frac{k}{n} \right)^2 \right) = \int_{0}^{1} \ln(1 + x^2) dx$.

Now, we need to evaluate the definite integral. This integral requires integration by parts. Let's set $u = \ln(1 + x^2)$ and $dv = dx$. Then, $du = \frac{2x}{1 + x^2} dx$ and $v = x$. Applying integration by parts, we get:

$\int_{0}^{1} \ln(1 + x^2) dx = x \ln(1 + x^2) \Big|_0^1 - \int_{0}^{1} \frac{2x^2}{1 + x^2} dx$.

Evaluating the first term, we have:

$x \ln(1 + x^2) \Big|_0^1 = 1 \cdot \ln(1 + 1^2) - 0 \cdot \ln(1 + 0^2) = \ln(2)$.

Now, we need to evaluate the integral $\int_{0}^{1} \frac{2x^2}{1 + x^2} dx$. We can rewrite the integrand as:

$\frac{2x^2}{1 + x^2} = 2 \frac{x^2}{1 + x^2} = 2 \frac{1 + x^2 - 1}{1 + x^2} = 2 \left(1 - \frac{1}{1 + x^2} \right)$.

Therefore, the integral becomes:

$\int_{0}^{1} \frac{2x^2}{1 + x^2} dx = 2 \int_{0}^{1} \left(1 - \frac{1}{1 + x^2} \right) dx = 2 \int_{0}^{1} dx - 2 \int_{0}^{1} \frac{1}{1 + x^2} dx = 2x \Big|_0^1 - 2 \arctan(x) \Big|_0^1 = 2 - 2 \cdot \frac{\pi}{4} = 2 - \frac{\pi}{2}$.

Substituting this back into the integration by parts equation, we get:

$\int_{0}^{1} \ln(1 + x^2) dx = \ln(2) - \left(2 - \frac{\pi}{2} \right) = \ln(2) - 2 + \frac{\pi}{2}$.

Thus, we have:

$\lim_{n \to +\infty} \ln(u_n) = \ln(2) - 2 + \frac{\pi}{2}$.

To find the limit of $u_n$, we need to exponentiate both sides:

$\lim_{n \to +\infty} u_n = e^{\ln(2) - 2 + \frac{\pi}{2}} = e^{\ln(2)} \cdot e^{-2} \cdot e^{\frac{\pi}{2}} = 2e^{\frac{\pi}{2} - 2}$.

Therefore, the limit of the sequence $u_n$ as $n$ approaches positive infinity is $2e^{\frac{\pi}{2} - 2}$. This result showcases the power of logarithmic transformations, Riemann sums, and integration by parts in tackling complex limit problems.

In summary, by transforming the product into a summation using logarithms, recognizing the Riemann sum structure, and employing integration by parts, we successfully determined the limit of the sequence $u_n$ as $n$ approaches infinity. The process involved a blend of algebraic manipulation, integral calculus, and limit evaluation, demonstrating the interconnectedness of mathematical concepts.

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In conclusion, the discussion category for this article is undoubtedly mathematics, encompassing areas such as calculus, analysis, sequences, limits, and integration. The problems serve as valuable exercises for students and enthusiasts alike, reinforcing fundamental concepts and showcasing the power and beauty of mathematical reasoning.