1. ## Existence of limit

Hello all,

I have the following problem: suppose that $\displaystyle \{S_{n} \}$ is a convergent sequence. Then we are asked to prove that $\displaystyle \lim_{n\rightarrow \infty}2S_{n}$ exists.

I have done the following:

By assumption I know that $\displaystyle \{S_{n} \}$ converges. This means that $\displaystyle \forall \epsilon>0$ we can find an $\displaystyle N\in \mathbb{N}$ such that $\displaystyle \mid S_{n}-L \mid<\epsilon$ whenever $\displaystyle n\geq N$.
We therefore have:
$\displaystyle \mid S_{n}-L \mid<\epsilon$
$\displaystyle 2\cdot\mid S_{n}-L \mid<2\epsilon$
$\displaystyle \mid 2S_{n}-2L \mid<2\epsilon$
$\displaystyle \mid 2S_{n}-2L \mid-\epsilon<2\epsilon-\epsilon$
$\displaystyle \mid 2S_{n}-2L \mid-\epsilon<\epsilon$

Could someone point in the right direction?

2. ## Re: Existence of limit

Originally Posted by detalosi
I have the following problem: suppose that $\displaystyle \{S_{n} \}$ is a convergent sequence. Then we are asked to prove that $\displaystyle \lim_{n\rightarrow \infty}2S_{n}$ exists.
Could someone point in the right direction?
We are given that $(s_n)\to L$
If $\varepsilon>0$ then $\dfrac{\varepsilon}{2}>0$ so $\exists N\in \mathbb{Z}^+$ such that $n > N \Rightarrow |{s_n} - L| < \dfrac{\varepsilon}{2}$
$|2{s_n} - 2L|=|s_n-L|(2) <(2) \dfrac{\varepsilon}{2}=\varepsilon$ By definition $(2s_n)\to 2L$

3. ## Re: Existence of limit

Thanks.

I have written it in the following form:

Since $\displaystyle \{ S_{n} \}$ converges to $\displaystyle L$ then $\displaystyle \forall \epsilon >0$ there is an $\displaystyle N\in \mathbb{N}$ such that $\displaystyle \mid S_{n}-L \mid <\epsilon$ whenever $\displaystyle n\geq N$. Since this is true for all $\displaystyle \epsilon>0$ then it is also true for $\displaystyle \frac{\epsilon}{2}>0$.
Thus we have:

$\displaystyle \mid S_{n}-L\mid<\frac{\epsilon}{2}$
$\displaystyle 2\cdot\mid S_{n}-L\mid<2\cdot\frac{\epsilon}{2}$
$\displaystyle \mid 2S_{n}-2L\mid<\epsilon$

By definition $\displaystyle 2S_{n}$ converges to $\displaystyle 2L$ and therefore the limit $\displaystyle \lim_{n\to \infty}2S_{n}$ exists.