Suppose $a_n$ is a convergent sequence. Then $|a_n-l|<\frac{1}{n}$?

Question

If a sequence of real numbers is convergent then is it true that $ \exists k \in \mathbb{N} $ so that $|a_n- l|<\dfrac{1}{n}$ for all $n\geq k$? If not then are there any sufficient conditions for this to hold? $l$ denotes the limit of the sequence.

Answer 1

In general, you have that $\exists N \in \mathbb{N}$ s.t. $\forall m> N$ we have $$|a_m-l|<\frac{1}{n}$$

which is the definition of convergence.

However, note that the $n$ in $\frac{1}{n}$ is different from the $N$ in the definition, hence your claim fails to hold and an example is the sequence $(a_n)$ defined by $a_n=\frac{2}{n}$. It converges to $0$ but $|\frac{2}{n}-0|>\frac{1}{n}$

Answer 2

Correct me if wrong:

As noted in the above answers the above statement is not true in general.

A little detour:

Given the convergent sequence

$(a_n)_{n \in \mathbb{Z^+}}$:

consider a sequence $(n_k)_{k \in \mathbb {Z^+}}$ of positive integers

$n_1< n_2< n_3<........$

then $(a_{n_k})$ is a subsequence .

Now:

Construct inductively a subsequence:

$k=1$: Choose $n_1$ such that for $n\ge n_1$:

$|a_n -L| \lt 1/1$.

$k=2$: Choose $n_2$ such that for $n\ge n_2$:

$| a_n - L| \lt 1/2$ .

......

.......

$k=m$:

Choose $n_m$ such that for $n \ge n_m$:

$|a_n-L| \lt 1/m$.

For this subsequence $(a_{n_k})$ we have:

$|a_{n_k} -L| < 1/k$, for $k =1,2,3....$