For a sequence of positive real numbers converging to a limit (which is not equal to 0), show that infimum > 0

Question

I think to do this I either need to prove the infimum is a minumum or a limit but I'm not sure how.

I have tried this:

By definition of the infimum: " Let A be a subset of the real numbers and b be a real number. Then b is the infimum of A if: (i) b≤a for all a ∈ A (ii) c≤a for all a ∈ A will imply that c≤b."

We know for all a ∈ A, a>0. If we consider the second case, this means the infimum (b) is such that 0≤b.

We can assume for a contradiction that b = 0 in order to leave just the case 0 < b. b cannot be the minimum of the set, as all a > 0. because all the terms are positive, for the infimum to be 0, it means the limit of the series is zero (and the sequence must be decreasing), but L cannot be equal to 0 so we have a contradiction.

This highlighted bit is the jump I'm struggling with.

Answer 1

Hint: you can use the fact the sequence converges to say the inf of all the numbers past some point $N$ is greater than, say, half the limit. Just get $N$ from whoever told you the sequence converges by giving them $\epsilon$ as half the limit. Now there are only finitely many numbers out to $N$.

Answer 2

Outline of what I'd try:

The limit can't be negative, since then terms approaching it would be negative. Its not allowed to be zero, so is positive.

Call the sequence $(x_n)$ and the limit $k$. Choose an $\epsilon$ such that $k-\epsilon >0$.

Beyond some $n=N$, all $x_n>k-\epsilon$ because of the convergence.

The first $N$ terms are all positive by definition, and have a minimum value $m$.

If the infimum is below both $m$ and $k-\epsilon$ it can't meet the definition of an infimum. So it's $\ge$ the lower of them, which is positive. Therefore the infimum is positive.

I think that does it, once it's translated into more precise language. (My aim here is to make it as human-readable as I can, to make clear how it works.)

Answer 3

Let $(x_n)$ be a sequence of positive real numbers such that $ \ \displaystyle \lim_{n \to \infty} x_n = L \neq 0 \ $. As $ \ \frac{|L|}{2}>0 \ $, by definition, there exists $ \ N \in \mathbb{N} \ $ such that, for all $ \ n \in \mathbb{N} \ $, if $ \ n>N \ $, then $ \ |x_n - L|< \frac{|L|}{2} \ $.

Let $ \ A = \{ 0,1,2,3,...,N \} \subset \mathbb{N} \ $, $ \ B = \mathbb{N} \setminus A = \{ N+1, N+2, N+3, ... \} \subset \mathbb{N} \ $ and $$m= \min \{ x_0 , x_1, x_2 , x_3 , ... , x_N \} \ .$$ Then $ \ \mathbb{N} = A \cup B \ $ and $ \ A \cap B = \varnothing \ $. Hence, for all $ \ n \in \mathbb{N} \, $, we have either $ \ $ (i) $ \ n \in A \ $ or $ \ $ (ii) $ \ n \in B \ $, for which

(i) $ \ $ if $ \ n \in A \ $, then $ \ 0 < m \leq x_n \ $ ;

and

(ii) $ \ $ if $ \ n \in B \ $, then $ \ n \in \mathbb{N} \ $ and $ \ n >N \ $ and we have that $ \ |x_n - L|< \frac{|L|}{2} \ $. Therefore $$|L| -|x_n| \leq \big| |L| - |x_n| \big| \leq |L - x_n| = |x_n - L| < \frac{|L|}{2} \ , $$ that is, $$x_n = |x_n|>|L|- \frac{|L|}{2}= \frac{|L|}{2} >0 \ . $$

Choosing $ \ c = \min \left\{ m , \frac{|L|}{2} \right\} \ $, we get that $ \ x_n \geq c>0 \ $, $\forall n \in \mathbb{N}$. So $ \ c>0 \ $ and $c$ is a lower bound for $(x_n)$. We conclude that $ \ \displaystyle \inf_{n \in \mathbb{N}} x_n \geq c > 0 \ $.

Answer 4

When studying new mathematical material, an attempt must be made be to write down 'the setup' in a mathematically precise way, and to avoid using logical arguments that use the informal language of, say, English.

The setup:

$\tag 1 \lim\limits_{n \to +∞} a_n = L, \text{ with } L \gt 0 \text{ and } a_n \gt 0$

$\tag 2 A = \{ a_n \, | \, n \in \mathbb N\}$

$\tag 3 b = \text{inf}(A)$

Assume $b = 0$.

Using open intervals of the form $(0, \varepsilon)$, a strictly decreasing sequence $b_n$ with $b_n \in A$ can be constructed such that

$\tag 4 \lim\limits_{n \to +∞} b_n = 0$

Now $b_n$ might not be a subsequence of $a_n$, but you can take another pass over it, throwing out terms if necessary to get the desired subsequence. Since every subsequence of $(a_n)_{ \, n \in \mathbb N} $must have the same limit, $L = 0$, we arrive at a contradiction.

Note that you can construct $b_n$ as a subsequence on the first pass, but thought it might be easier to follow by breaking it into two steps. The basic truth is that for any finite set of positive numbers, an $\varepsilon \gt 0$ can be found such that the open interval $(0, \varepsilon)$ is disjoint from that set.

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The OP states in bold

it means the limit of the series is zero (and the sequence must be decreasing)

Let $(a_n)_{ \, n \in \mathbb N} $ be defined by

$$ a_n = \left\{\begin{array}{lr} \frac{1}{n+1}, & \text{for } n \text{ even }\\ \frac{1}{n-.5}, & \text{for } n \text{ odd } \end{array}\right\} $$

This sequence of positive numbers converges to $0$, but it is not a decreasing sequence
(see the wikipedia definition).