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State whatever the sequence converges and, if it does, find the limit. $$\sqrt{n^{2}+n}-n$$

Short Answer

Expert verified
The sequence \(\sqrt{n^2 + n} - n\) converges, and its limit is \(\frac{1}{2}\).

Step by step solution

01

Identify the expression

The given sequence is \(\sqrt{n^2 + n} - n\), and our task is to investigate its limit as n approaches infinity.
02

Simplify the expression

We can factor out n from the expression inside the square root. The expression becomes: \(\sqrt{n^2 + n} - n = \sqrt{n(n + 1)} - n\) Now, let's try to further simplify the expression by using a common trick: multiplying and dividing by the conjugate of the expression. Let the conjugate be \(\sqrt{n(n+1)} + n\), then we get: \(\frac{\sqrt{n(n + 1)} - n}{1} \cdot \frac{\sqrt{n(n+1)} + n}{\sqrt{n(n+1)} + n}\)
03

Carry out the multiplication

Now, we need to multiply the two expressions term by term: \(\frac{(\sqrt{n(n+1)} - n)(\sqrt{n(n+1)} + n)}{1(\sqrt{n(n+1)} + n)}\) When we multiply the numerator, we are left with the difference of two squares, so we obtain: \(\frac{n(n+1) - n^2}{\sqrt{n(n+1)} + n}\)
04

Simplify the fraction

Now, we can simplify the numerator by expanding and subtracting the terms: \(\frac{n^2 + n - n^2}{\sqrt{n(n+1)} + n} = \frac{n}{\sqrt{n(n+1)} + n}\) Now, let's factor n out of the numerator and the denominator: \(\frac{n}{\sqrt{n(n+1)} + n} = \frac{n}{n(\frac{\sqrt{n+1}}{\sqrt{n}} + 1)}\)
05

Cancel out common factors

Now we can cancel out the common factor n from the numerator and denominator: \(\frac{n}{n(\frac{\sqrt{n+1}}{\sqrt{n}} + 1)} = \frac{1}{\frac{\sqrt{n+1}}{\sqrt{n}} + 1}\)
06

Find the limit

Now, we need to find the limit of the simplified expression as n approaches infinity: \(\lim_{n \to \infty} \frac{1}{\frac{\sqrt{n+1}}{\sqrt{n}} + 1}\) As n approaches infinity, the term inside the square root becomes relatively small compared to n: \(\frac{\sqrt{n+1}}{\sqrt{n}} \approx \frac{\sqrt{n}}{\sqrt{n}} = 1\) So, the limit is: \(\lim_{n \to \infty} \frac{1}{1+1} = \frac{1}{2}\) The given sequence converges, and its limit is \(\frac{1}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergent Sequence
Imagine you're on a journey towards a fixed destination, but with each step, you halve the distance between yourself and the goal. Theoretically, you are ever closer but never quite there, yet for practical purposes, we can say you'll eventually arrive. A similar idea applies to a convergent sequence in mathematics.

When a sequence's terms get closer and closer to a specific value as you move through the sequence, it's called a convergent sequence. The fixed destination is akin to the sequence's limit. For instance, the given problem involves the sequence \(\sqrt{n^{2}+n}-n\). By applying algebraic techniques and understanding the behavior of the sequence as \(n\) grows larger, we determine that the destination, or the limit, is \(\frac{1}{2}\). Convergence essentially tells us where the sequence's journey ends after an infinite number of steps.
Infinite Limits
The concept of infinite limits guides us through scenarios where our function's values either increase without bound or decrease without bound as we approach a certain point. This is distinctly different from what we see in convergent sequences, which approach a specific number. We're not talking about the limit being infinitely large, but rather considering what happens as the variable, in this case \(n\), heads towards infinity.

In the given sequence, we explore the limit as \(n\) grows indefinitely - a journey into the infinite. But instead of the values blowing up to infinity, they settle down, approaching the finite value of \(\frac{1}{2}\). This is a beautiful illustration that even as we venture towards infinity with our variable, the sequence itself may still comfortably converge to a finite limit.
Difference of Squares
The difference of squares is a nifty algebraic pattern that surfaces quite frequently in mathematical problems. It is when you have two terms squared and subtracted from one another, represented as \(a^2 - b^2\). This special form allows us to factor the expression into \(a + b)(a - b)\), which is exactly what's employed in the solution to our original exercise.

By multiplying the sequence's terms \(\sqrt{n(n+1)} - n\) by its conjugate \(\sqrt{n(n+1)} + n\), we use the difference of squares idea to simplify the expression. The result? A cleaner and more manageable form that ultimately leads us to the convergence point. Spotting a difference of squares in the wild can be a crucial step to unraveling more complex problems, just like this one!
Limit of a Function
The limit of a function is a foundational concept within calculus that describes the behavior of a function as the input approaches a particular value. It doesn't have to be about what the function equals at that point, but where it's 'headed'.

In our exercise, we investigate the limit as \(n\) approaches infinity, which tells us about the long-term behavior of the sequence \(\sqrt{n^{2}+n}-n\). The limit concept is seamlessly woven into the exercise, as we break down and simplify the expression, and analyze how the terms behave in the grand scale of growing numbers. This grasp on the concept of limits is essential not just for single problems, but for an entire array of calculus-related topics.

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Most popular questions from this chapter

Laplace transforms. Let \(f\) b: continuous on \([0,\infty)\). The Laplace transform of \(f\) is the function \(F\) defined by setting $$F(s)=\int_{0}^{\infty} e^{-s x} f(x) d x$$ The domain of \(F\) is the set of numbers \(s\) for which the improper integral converges. Find the Laplace transform \(F\) of each of the following functions specifying the domain of \(F\). $$f(x)=1$$

Let \(\Omega\) be the region bounded below by \(y\left(x^{2}+1\right)=x,\) above by \(x y=1,\) and to the left by \(x=1\). (a) Find the area of \(\Omega\) (b) Show that the configuration obtained by revolving \(\Omega\) about the \(x\) -axis has finite volume. (c) Calculate the volume generated by revolving \(\Omega\) about the \(y\) -axis.

Give the first six terms of the sequence and then give the \(n\) th term. $$a_{1}=1 ; \quad a_{n+1}=\frac{n}{n+1} a_{n}$$.

Let \(S=\left\\{a_{1}, a_{2}, a_{3}, \cdots, a_{n}, \cdots\right\\}\) with \(a_{1}=4\) and for further subscripts \(a_{n+1}=3-3 / a_{n}.\) (a) Calculale the numbers \(a_{2}, a_{3} . a_{4}, \cdots, a_{10}\). (b) Use a graphing utility or CAS to calculate \(a_{20}\) \(a_{30}, \cdots, a_{50}\). (c) Does \(S\) have a least upper bound? If so, what is it? Does S have a greatest lower bound? If so, what is it?

Starting with \(0 < a < b,\) form the arithmetic mean \(a_{1}=\frac{1}{2}(a+b)\) and the geometric mean \(b_{1}=\sqrt{a b} .\) For \(n=2.3,4, \cdots\) set $$a_{n}=\frac{1}{2}\left(a_{n-1}+b_{n-1}\right) \quad and \quad b_{n}=\sqrt{a_{n-1} b_{n-1}}$$ (b) Show that the two sequences converge and \(\lim _{n \rightarrow \infty} a_{n}=\) \(\lim _{x \rightarrow \infty} b_{n} .\) The common value of this limit is called the aruhmetic-geontetric mean of a and b.

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