Question

Suppose {an} is a sequence that converges to a number
L > 0.

What is the limit of {an + 1}?

Answer 1

Since it converges, you have that for all \(\epsilon>0\), \(|L-a_n|<\epsilon\) for all \(n>N\) for some \(N\in\mathbb{N}\).

Now, we guess that \(\{a_n+1\}\) converges to \(L+1\). Look at \(|L+1-(a_n+1)|\). What can you say about it?

Answer 2

It's a with subscript (n+1) . I'm ask confused with your symbols

Answer 3

Ah. So your new sequence is \(\{a_{n+1}\}\)?

Answer 4

Yes

Answer 5

Hey, @kimmy0394
When you think about it
\[\huge a_{n+1}\]is just the same sequence, just that it starts one step ahead of \[\huge a_n\]
right?

Answer 6

right!