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Chapter 3: Problem 12

Show that \(\lim \left(1 / 3^{n}\right)=0\).

Short Answer

Expert verified
\(\lim_{n \rightarrow \infty} \frac{1}{3^n} = 0\)

Step by step solution

01

Understanding the problem

Given is a sequence \(a_n = \frac{1}{3^n}\), which decreases as n increases. We are asked to verify that \(\lim_{n \to \infty} \frac{1}{3^n} = 0\)
02

Checking the sequence for n=1,2,3...

We first check the sequence by substituting n = 1, 2, 3, etc. into the formula \(a_n = \frac{1}{3^n}\) to see if the sequence values are getting closer to 0.
03

Checking the limit

The limit of a sequence as n approaches infinity is defined as \(\lim_{n \to \infty} a_n\). Substituting the given sequence into this formula we get \(\lim_{n \to \infty} \frac{1}{3^n}\)
04

Observing the sequence

It can be observed that as n increases, the term \(\frac{1}{3^n}\) gets closer and closer to 0. This is because the denominator \(3^n\) grows much faster than the numerator 1, hence the fraction becomes increasingly smaller.
05

Concluding the limit

As when \(n \rightarrow \infty\), the value of the sequence \(\frac{1}{3^n}\) approaches 0, we can hence conclude that \(\lim_{n \rightarrow \infty} \frac{1}{3^n} = 0\)

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Most popular questions from this chapter

If \(x_{1}2\), show that \(\left(x_{n}\right)\) is convergent. What is its limit?

Let \(y_{1}:=\sqrt{p}\), where. \(p>0\), and \(y_{n+1}:=\sqrt{p+y_{n}}\) for \(n \in \mathbb{N}\). Show that \(\left(y_{n}\right)\) converges and find the limit. [Hint: One upper bound is \(1+2 \sqrt{p} .]\)

Show that the following sequences are not convergent. (a) \(\left(2^{n}\right)\), (b) \(\left((-1)^{n} n^{2}\right)\).

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum \sqrt{a_{n} a_{n+1}}\) always convergent? Either prove it or give a counterexample.

For any \(b \in \mathbb{R}\), prove that \(\lim (b / n)=0\).
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