# prove that if sequence (an) converges, use def of limit of sequence to prove...

• Aug 28th 2009, 02:34 AM
serenalee
prove that if sequence (an) converges, use def of limit of sequence to prove...
that if an>=1 for all natural number n, then L >= 1.

From the def of limit of sequence,

L-a(n) < epsilon and a(n) - L < epsilon

we know that a(n) >=1,

L - a(n) < epsilon => L must be larger than or same with a(n) to keep epsilon positive.

a(n) - L < epsilon => L must be larger than or same with a(n) to keep the equation on the left smaller than epsilon.

Is this argument acceptable?
• Aug 28th 2009, 02:56 AM
Defunkt
Quote:

Originally Posted by serenalee
that if an>=1 for all natural number n, then L >= 1.

From the def of limit of sequence,

L-a(n) < epsilon and a(n) - L < epsilon

we know that a(n) >=1,

L - a(n) < epsilon => L must be larger than or same with a(n) to keep epsilon positive.

a(n) - L < epsilon => L must be larger than or same with a(n) to keep the equation on the left smaller than epsilon.

Is this argument acceptable?

This is incorrect.

In your proof you say that because $L - a_n < \epsilon$ , $L \geq a_n$ for $\epsilon$ to be positive, however if $L < a_n$ then $L - a_n < 0 \Rightarrow L - a_n < \epsilon$ for all $\epsilon > 0$

My idea of proving this would be to assume, in contradiction, that L < 1 and then you can find some $\epsilon$ that has no N such that for all n>N $|a_n-L|<\epsilon$. Can you follow from here?

Spoiler:
Proof:
Assume by contradiction that $L<1$ and let $\epsilon_0 = 1-L$ . Assume that there exists $n_0 \in \mathbb{N}$ such that for all $n>n_0 , |a_n - L| < \epsilon_0$
Then, $|a_n - L| < 1-L \Rightarrow a_n - L < 1 - L \Rightarrow a_n<1$ - contradiction.
• Aug 28th 2009, 02:59 AM
Taluivren
Hi
the definition of limit says that for each $\varepsilon > 0$ there exists $n_0 \in \mathbb{N}$ such that for each $n > n_0$ we have $\left|L-a_n\right| < \varepsilon$.
Let's argue by contradiction: assume that $L < 1$. Set $\varepsilon = \frac{1-L}{2}$. Then, according to the definition, there must exist $n_0 \in \mathbb{N}$ such that for each $n > n_0$ we have $\left|L-a_n\right| < \varepsilon = \frac{1-L}{2}$. But this means $a_n < 1$, a contradiction.