Question

Evaluate the following limits: (i) \(\lim _{[\rightarrow+} \sum_{t=1}^{n} \frac{n}{(n+r) \sqrt{r(2 n+r)}}\) (ii) \(\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{\sqrt{n^{2}-r^{2}}}\) (iii) \(\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n+r}\)

Short answer

#Problem# (i) Evaluate the following limit: $$\lim _{n \rightarrow+} \sum_{t=1}^{n} \frac{n}{(n+r) \sqrt{r(2 n+r)}}$$ (ii) Evaluate the following limit: $$\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1}\frac{1}{\sqrt{n^{2}-r^{2}}}$$ (iii) Evaluate the following limit: $$\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n+r}$$ #Solution# (i) The evaluated limit for the given summation is: $$\sum_{k=n+1}^{2n}\frac{n}{k\sqrt{(k-n)(3n - k)}}$$ (ii) The evaluated limit for the given summation is: $$\frac{\pi}{2}$$ (iii) The evaluated limit for the given summation is: $$\ln(3)$$

Step-by-step solution

Rewrite the problem

The problem can be written as: \(\lim _{[\rightarrow+} \sum_{t=1}^{n} \frac{n}{(n+r) \sqrt{r(2 n+r)}}\) = \(\lim _{[\rightarrow+} \sum_{t=1}^{n} \frac{n}{(n+r) \sqrt{r(2 n+r)}}\)

Introduce a substitution

Let's start by making a substitution: let \(k = n + r\), then our sum changes to \(\sum_{k=n+1}^{2n}\frac{n}{k\sqrt{(k-n)(2n+n - k)}}\)

Evaluate the sum

Now let's sum the problem: \(\sum_{k=n+1}^{2n}\frac{n}{k\sqrt{(k-n)(3n - k)}}\) (ii)

Rewrite the problem

The problem can be written as: \(\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1}\frac{1}{\sqrt{n^{2}-r^{2}}}\)

Identify the Riemann sum

Observe that the given expression can be thought of as a Riemann sum. More specifically, \(\frac{1}{n} \sum_{r=0}^{n-1}\frac{1}{\sqrt{1-(\frac{r}{n})^2}}\)

Find the continuous function

From the previous step, we identify the continuous function to be \(f(x)=\frac{1}{\sqrt{1-x^2}}\) on the interval \([0,1]\).

Evaluate the integral

Now, we have the limit to be: \(\lim _{n \rightarrow \infty} \left(\frac{1}{n}\sum_{r=0}^{n-1}f(\frac{r}{n})\right)\) By definition of Riemann sum, this evaluates to the integral of \(f(x)\) from 0 to 1: \(\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\) (iii)

Rewrite the problem

The problem can be written as: \(\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n+r}\)

Factor out n

We can factor out the n from the denominator: \(\sum_{r=1}^{2 n}\frac{1}{n(1+\frac{r}{n})}\)

Identify the Riemann sum

Observe that the given expression can be thought of a Riemann sum. Specifically, \(\frac{1}{n} \sum_{r=1}^{2n}\frac{1}{1+(\frac{r}{n})}\)

Find the continuous function

From the previous step, we identify the continuous function to be \(g(x) = \frac{1}{1+x}\) on the interval \([0,2]\).

Evaluate the integral

Now, we have the limit to be: \(\lim _{n \rightarrow \infty} \left(\frac{1}{n}\sum_{r=1}^{2n}g(\frac{r}{n})\right)\) By definition of Riemann sum, we evaluate the integral of \(g(x)\) from 0 to 2: \(\int_{0}^{2}\frac{1}{1+x}dx=\ln(3)\)

Key concepts

Riemann Sum

Riemann sums are a fundamental concept in calculus, used as a method for approximating the area under a curve—essentially measuring the integral of a function. Imagine you're trying to estimate the amount of paint needed for the hilly landscape of a mural. By breaking the landscape into a sequence of vertical slices and then calculating the area of each slice, you're essentially performing what mathematicians call a Riemann sum.

In the context of an exercise like the second one above, a Riemann sum takes the form of adding up areas of rectangles under a curve where the width of each rectangle is vanishingly small, and the function's value at a chosen point within each interval serves as the height. This process can be symbolized as \( \frac{1}{n} \sum_{r=0}^{n-1}f(\frac{r}{n}) \) for some function \(f\), with \(1/n\) representing the width of each rectangle becoming smaller as \(n\) grows. As \(n \rightarrow \infty\), these sums approach the exact area under the curve—this is what we call the definite integral of the function over the interval.

Integral Calculus

Integral calculus is the reverse process of differentiation and serves as a tool for finding quantities like areas, volumes, and sums. It's like piecing together the growth history of a plant by looking at each segment. The central tool of integral calculus is the definite integral, denoted by \( \int_{a}^{b} f(x)dx \)—it represents the sum of infinitesimal slices \(dx\) multiplied by the function's value at each slice \(f(x)\).

When we calculated the limit in the second part of the exercise, it ultimately translated into finding the integral \( \int_{0}^{1}\frac{1}{\sqrt{1-x^2}}dx \) because the sum represented the area under the curve. The result, \(\frac{\pi}{2}\), gives the exact area of a quarter-circle, which in geometric terms, relates directly back to the infinite series we're summing to evaluate our Riemann sum.

Infinite Series

An infinite series is like a never-ending path made up of stepping stones; each stone is a term in the series, and the sum is the total distance you'd cover if you could step on every stone infinitely. Mathematically, an infinite series adds up an infinite number of terms to possibly arrive at a finite value.

In our third exercise example, the infinite series is represented as \( \lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n+r}\), which evaluates to \( \ln(3)\) upon using integral calculus. This process connects to integral calculus because, in a sense, finding the sum of an infinite series often involves taking the limit of Riemann sums—as \(n\) increases without bound, we find the 'sum' of an infinitely fine partition, which can be interpreted as the area under a curve, thus linking back to the integral of a function.