Is sequence 1/(n-1) necessarily bounded?

[Axel Harper]

Homework Statement

Courant states that a convergent sequence is necessarily bounded; that is, for all n, the absolute value of term a_n is less than or equal to some number M. My question is does this apply to the sequence given by a_n = 1/(n-1)?

Homework Equations

As n approaches infinity, a_n approaches zero, so the sequence converges.

The Attempt at a Solution

At n = 1, 1/(n-1) is larger than any number M, which suggests the sequence is not bounded. Should we only consider terms for n > 1, in which case the sequence would be bounded by 1 and still converge to 0?

 

[Mark44]

Homework Statement

Courant states that a convergent sequence is necessarily bounded; that is, for all n, the absolute value of term a_n is less than or equal to some number M. My question is does this apply to the sequence given by a_n = 1/(n-1)?

Homework Equations

As n approaches infinity, a_n approaches zero, so the sequence converges.

The Attempt at a Solution

At n = 1, 1/(n-1) is larger than any number M, which suggests the sequence is not bounded. Should we only consider terms for n > 1, in which case the sequence would be bounded by 1 and still converge to 0?

Since a₁ is undefined, the sequence makes sense only for n >= 2.

 

[vela]

A sequence is a function. Rather than writing f(n), however, we use the convenient notation $a_n$. In the example you've given, n=1 isn't in the domain of the function, so there is no $a_1$ in that sequence. So, yes, you only consider the terms for n>1.

Also, I wouldn't say that at n=1, 1/(n-1) is larger than any number M because 1/(n-1) is not defined for n=1. If it's not defined, it doesn't make sense to talk about whether it's larger or smaller than some real number M.

What you're thinking of is the real function $f: \mathbb{R}-\{1\} \to \mathbb{R}$ where f(x) = 1/(x-1). As before f(1) isn't defined, and it doesn't make sense to say f(1) is smaller or bigger than some number. Consider the fact that as x approaches 1, the function diverges to $\pm\infty$ depending on which side you approach x=1 from, so is f(1) really big or really small?

 

[Fredrik]

Homework Statement

Courant states that a convergent sequence is necessarily bounded; that is, for all n, the absolute value of term a_n is less than or equal to some number M. My question is does this apply to the sequence given by a_n = 1/(n-1)?

Homework Equations

As n approaches infinity, a_n approaches zero, so the sequence converges.

The Attempt at a Solution

At n = 1, 1/(n-1) is larger than any number M, which suggests the sequence is not bounded. Should we only consider terms for n > 1, in which case the sequence would be bounded by 1 and still converge to 0?

The sequence $\big(\frac{1}{n-1}\big)_{n=2}^\infty$ is convergent, and its limit is 0, as you said. But there's no sequence $\big(\frac{1}{n-1}\big)_{n=1}^\infty$, because $\frac 1 0$ is not larger than any number. It's just undefined.

A convergent sequence is bounded because "convergent" means that for all $\varepsilon>0$ there's a number x such that all but a finite number of terms are in the interval $(x-\varepsilon,x+\varepsilon)$. This makes it very easy to see that there's an $r>0$ such that all terms are in the interval $(x-r,x+r)$.

Edit: I wrote this about 45 minutes ago, but got distracted by something before I could submit it.