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Chapter 3: Problem 11

Show that \(\lim \left(\frac{1}{n}-\frac{1}{n+1}\right)=0\).

Short Answer

Expert verified
The limit of the expression as \(n\) tends to infinity is \(0\).

Step by step solution

01

Simplify the expression

The expression can be simplified by finding a common denominator of \(n(n+1)\), which when utilized gives \(\frac{n+1-n}{n(n+1)}\).
02

Further Simplify the expression

The numerator of the expression simplifies to 1, which gives \(\frac{1}{n(n+1)}\).
03

Apply the limit

Now we are asked to find \(\lim_{n\to\infty}\frac{1}{n(n+1)}\). As \(n\) tends to infinity, both \(n\) and \(n+1\) tends to infinity. Thus the value of the term \(\frac{1}{n(n+1)}\) tends to \(0\).

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Key Concepts

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

Sequences
Sequences are a foundational concept in mathematics that represent ordered lists of numbers. Each number in this list is called a term. In our exercise, the sequence is given by the expression \( \frac{1}{n} - \frac{1}{n+1} \). As \( n \) increases, this sequence generates new terms that approach a specific behavior or value.

Understanding sequences helps us identify patterns or trends within numbers.
  • Each term in a sequence can be defined explicitly by a formula, like in our case, \( \frac{1}{n(n+1)} \).
  • The order of the terms is very important as it dictates how the sequence behaves over time.
Sequences are often used as a stepping stone to understanding more complex mathematical concepts, such as limits and convergence. As we observe the behavior of terms in a sequence, it becomes fundamental to limit analysis and deciding how terms behave as they grow large.
Infinity
Infinity is a concept used to describe something without any bounds or limits. In the context of sequences and limits, infinity often represents what happens as we move towards the very large or unbounded numbers.

In our solution, as \( n \) tends to infinity, \( n \) and \( n+1 \) both grow large without bound. This impacts our ability to directly compute values but allows us to discuss trends or limits.
  • Infinity is not a number, but a concept describing unbounded growth.
  • When we say \( n \to \infty \), it implies considering the behavior of something as \( n \) grows very large.
Understanding infinity is crucial when evaluating limits because it allows us to talk about what terms approach or "tend" to as indexes grow very large. In our example, the expression \( \frac{1}{n(n+1)} \) tends towards zero as \( n \) becomes infinitely large.
Convergence
Convergence in sequences refers to the idea that as the sequence progresses, the terms get closer to a specific value - called the limit. A sequence is said to converge if its terms approach a particular constant value as the index (in this case, \( n \)) increases.

In our task, we are asked to show that the limit of the sequence \( \frac{1}{n(n+1)} \) as \( n \to \infty \) is zero, which means this sequence converges to 0.
  • If the sequence approaches a limit, no matter how small the difference between the sequence terms and the limit is, we eventually reach a point where all subsequent terms are within this difference of the limit.
  • A converging sequence has a predictability embedded within, making them extremely useful for calculations and mathematical proofs.
Convergence is pivotal in mathematical analysis, as it informs us about the stability and predictable behavior of sequences as their terms progress towards infinity.

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Most popular questions from this chapter

Suppose that \(\left(x_{n}\right)\) is a convergent sequence and \(\left(y_{n}\right)\) is such that for any \(\varepsilon>0\) there exists \(M\) such that \(\left|x_{n}-y_{n}\right|<\varepsilon\) for all \(n \geq M\). Does it follow that \(\left(y_{n}\right)\) is convergent?

Let \(\left(x_{n}\right)\) be properly divergent and let \(\left(y_{n}\right)\) be such that \(\lim \left(x_{n} y_{n}\right)\) belongs to \(\mathbb{R}\). Show that \(\left(y_{n}\right)\) converges to 0 .

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, and if \(b_{n}:=\left(a_{1}+\cdots+a_{n}\right) / n\) for \(n \in \mathbb{N}\), then show that \(\sum b_{n}\) is always divergent.

Show that if \(X\) and \(Y\) are sequences such that \(X\) converges to \(x \neq 0\) and \(X Y\) converges, then \(Y\) converges.

Is the sequence \((n \sin n)\) properly divergent?
See all solutions

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