4

Suppose $(a_n)$ is convergent, say $a_n \to L$. Then $a_{n+1} \to L$ as well.

Here is my confusion:

IS $a_{n+1}$ meant to be the $(n+1)st$ term of the sequence? or is it the subsequence $(a_2,a_3,a_4,....)$??

Reason of the question: In my book, the proof that if $\sum a_n $ converges , then $lim a_n = 0$ goes as follows:

Let $(s_n)$ be sequence of partial sums. Since $\sum a_n$ converges, then $s_n \to L$. Notice $s_{n+1} - s_n = a_n $ and

$$ \lim a_n = \lim (s_{n+1} - s_n) = s - s = 0$$

so in here they are asuming $s_{n+1} \to L$ as well, but why?

2 Answers2

4

Yes, $a_{n+1}$ is the $(n+1)$th term of the sequence, just like $a_n$ is the $n$th term.

To answer your other question, it's because if $n \to +\infty$ then $n+1 \to +\infty$ also. So then $$ \lim_{n\to+\infty} s_{n+1} = \lim_{n+1 \to+\infty} s_{n+1}.$$

Now just make the substitution $m=n+1$.

1

A more rigorous approach would be to note that as $\left \{ a_n \right \}_{n\in \mathbb N}$ converges to $L$, so does every subsequence, and $\left \{ a_{n+1} \right \}_{n\in \mathbb N}$ is one such. This can be easily seen as soon as we define $n:\mathbb N\to \mathbb N$ by $n(k)=k+1$, for then, $a_{n_{k}}=a_{k+1}.$

Matematleta
  • 29,139