Given $c_n \rightarrow L$ and $d_n$ lies between $c_n, c_{n+1}$, show $d_n \rightarrow L$.

Question

Would this make use of the squeeze theorem? That is, do I define two subsequences $\{c_n\}$ and $\{c_{n+1}\}$, show their limit is $L$ and then can conclude $d_n \rightarrow L$?

Answer 1

Typically, the squeeze theorem is stated as

If $x_n\to L$ and $y_n\to L$ and for all $n$, $x_n\le z_n\le y_n$, then $z_n\to L$

and not as

If $x_n\to L$ and $y_n\to L$ and for all $n$, $x_n\le z_n\le y_n$ or $x_n\ge z_n\ge y_n$, then $z_n\to L$

In the first form, the squeeze theorem won't help you - at least not naively: You may sometimes have $c_n\le d_n\le c_{n+1}$ and sometimes $c_n\ge d_n\ge c_{n+1}$. So if you do not have the second form available, you might try with $\min\{c_n,c_{n+1}\}\le d_n\le\max\{c_n,c_{n+1}\}$. Or just do it directly: If $|c_n-L|<\epsilon$ for all $n>N$, then ...

Answer 2

Mimic the proof of Squeeze Theorem: For $\epsilon>0$, choose an $N$ such that $L-\epsilon<c_{n}<L+\epsilon$ for $n\geq N$, then $d_{n}>\min\{c_{n},c_{n+1}\}>L-\epsilon$ and $d_{n}<\max\{c_{n},c_{n+1}\}<L+\epsilon$ and hence $|d_{n}-L|<\epsilon$ for all such $n$, this gives the result.