suppose that $lim_{n} a_{n}=l$ for some value of $l$, and that $a_{n}>0$ for $n$ even, and $a_{n}<0$ for $n$…
suppose that $lim_{n} a_{n}=l$ for some value of $l$, and that $a_{n}>0$ for $n$ even, and $a_{n}<0$ for $n$ odd. show that $l = 0$.
Answer
Explanation:
Step1: Use the definition of the limit
By the definition of the limit of a sequence, for any (\epsilon>0), there exists (N\in\mathbb{N}) such that for all (n > N), (|a_{n}-L|<\epsilon).
Step2: Consider even and odd (n) separately
- For even (n): Since (a_{n}>0) for even (n), we have (a_{n}-L<\epsilon) (because (|a_{n}-L| = a_{n}-L) when (a_{n}\geq L)). So (L>a_{n}-\epsilon). As (n\to\infty), taking the limit (since (\lim_{n\to\infty}a_{n}=L) for even (n) as well), and for any (\epsilon>0), (L\geq-\epsilon).
- For odd (n): Since (a_{n}<0) for odd (n), we have (L - a_{n}<\epsilon) (because (|a_{n}-L|=L - a_{n}) when (L\geq a_{n})). So (L<a_{n}+\epsilon). As (n\to\infty), taking the limit (since (\lim_{n\to\infty}a_{n}=L) for odd (n) as well), and for any (\epsilon>0), (L\leq\epsilon).
Step3: Conclude the value of (L)
Since for any (\epsilon>0), (-\epsilon\leq L\leq\epsilon), the only value that satisfies this for all (\epsilon>0) is (L = 0).
Answer:
(L = 0)