Question

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 $l = 0$.

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 \(L0\), \(L\leq\epsilon\).

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:

\(L = 0\)