Leibniz Criterion
A convergence criterion for sums based on an alternating sequence.
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Study Leibniz Criterion
For some particular cases, we nevertheless are able to show convergence.
Theorem 1 (Leibniz Convergence Criterion). Let \((a_n)_{n\in\mathbb{N}}\) be a monotonically decreasing real sequence with \[\lim_{n\to\infty}a_n=0.\] Then the alternating sequence \[\sum_{k=1}^\infty (-1)^ka_k\] converges.
Proof: Since \((a_n)_{n\in\mathbb{N}}\) is convergent to zero and monotonically decreasing, we have \(a_n\geq0\) for all \(n\in\mathbb{N}\). Let \((s_n)_{n\in\mathbb{N}}\) be the corresponding sequence of partial sums. Then we have for all \(l\in \mathbb{N}\) that \[s_{2l+2}-s_{2l}=-a_{2l+1}+a_{2l+2}\leq 0,\quad s_{2l+3}-s_{2l+1}=a_{2l+2}-a_{2l+3}\geq 0,\] i.e., the subsequence \((s_{2l})_{l\in\mathbb{N}}\) is monotonically decreasing and \((s_{2l+1})_{l\in\mathbb{N}}\) is monotonically increasing. Furthermore, due to \(s_{2l+1}-s_{2l}=-a_{2l+1}\leq0\) holds \[s_{2l+1}\leq s_{2l}.\]
Altogether, the \((s_{2l})_{l\in\mathbb{N}}\) is
monotonically decreasing and bounded from below, and \((s_{2l+1})_{l\in\mathbb{N}}\) is
monotonically increasing and bounded from above. By the Theorem on
monotonic and bounded sequences, both subsequences are convergent. Due
to \[\lim_{l\to\infty}(s_{2l+1}-s_{2l})=\lim_{l\to\infty}a_{2l+1}=0,\]
an application of the formulae for convergent sequences yields that both
subsequence have the same limit, i.e., \[\lim_{l\to\infty}s_{2l+1}=\lim_{l\to\infty}s_{2l}=s\]
for some \(s\in\mathbb{R}\). Let \(\varepsilon>0\). Then there exists some
\(N_1\) such that for all \(l\geq N_1\) holds \(|s_{2l}-s|<\varepsilon\). Furthermore,
there exists some \(N_2\) such that for
all \(l\geq N_2\) holds \(|s_{2l+1}-s|<\varepsilon\). Now choosing
\(N=\max\{2N_1,2N_2+1\}\), we can say
the following for some \(m\geq
N\):
In the case where \(m\) is even, we
have some \(l\in\mathbb{N}\) with \(m=2l\). By the choice of \(N\), we also have \(l\geq N_1\) and thus \[|s_{m}-s|=|s_{2l}-s|<\varepsilon.\] In
the case where \(m\) is odd, we have
some \(l\in\mathbb{N}\) with \(m=2l+1\). By the choice of \(N\), we also have \(l\geq N_2\) and thus \[|s_{m}-s|=|s_{2l+1}-s|<\varepsilon.\]
\(\Box\)
Example 2.
The alternating harmonic series \[\sum_{k=1}^\infty \frac{(-1)^k}{k}.\] converges. The limit is (without proof) \(\log(2)\).
The alternating Leibniz series \[\sum_{k=1}^\infty \frac{(-1)^k}{2k+1}.\] converges. The limit is (without proof) \(\frac\pi4\).
The series \[\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}.\] converges. The limit is not expressible in a closed form.
We will later treat the topic of Taylor series. Thereafter we will be able to determine some further limits of sequences.
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