Limit Theorems
Combining limits is a useful tool to deduce convergence and the limit of a more complicated sequence from the convergence of simpler building blocks.
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Study Limit Theorems
Theorem 1 (Formulae for convergent sequences). Let \((a_n)_{n\in\mathbb{N}}\) and \((b_n)_{n\in\mathbb{N}}\) be a convergent sequences in \(\mathbb{R}\). Then the following holds true:
\((a_n+b_n)_{n\in\mathbb{N}}\) is convergent with \[\lim_{n\to\infty}(a_n+b_n)= \lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n.\]
\(( a_n\cdot b_n)_{n\in\mathbb{N}}\) is convergent with \[\lim_{n\to\infty}(a_n\cdot b_n)= \lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}b_n.\]
If \(\lim_{n\to\infty}b_n\neq0\) and \(b_n\neq0\) for all \(n\in\mathbb{N}\), then the sequence \((\frac{a_n}{b_n})_{n\in\mathbb{N}}\) is convergent with \[\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{\displaystyle\lim_{n\to\infty}a_n}{\displaystyle\lim_{n\to\infty}b_n}.\]
Remark 2.
Since the constant sequence \((a)_{n\in\mathbb{N}}=(a,a,a,\ldots)\) is, of course, convergent to \(a\), statement (ii) also implies the formula \[\lim_{n\to\infty} ( a\cdot b_n ) = a\cdot\lim_{n\to\infty}b_n.\]
For \(k\in\mathbb{N}\), a \(k\)-times application of statement (ii) yields that for some convergent sequence \((a_n)_{n\in\mathbb{N}}\), also the sequence \((a_n^k)_{n\in\mathbb{N}}\) is convergent with \[\lim_{n\to\infty}a_n^k=\left(\lim_{n\to\infty}a_n\right)^k.\]
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