Notes - Analysis I MT22, Alternating series test


Flashcards

What’s the statement of the alternating series test for $\sum _ {k=1}^\infty (-1)^{k+1}a _ k$?


If $(a _ k)$ is a monotonically decreasing sequence and $a _ k \to 0$, then $\sum^\infty _ {k=1}(-1)^{k+1}a _ k$ converges.

What’s the basic idea when proving the alternating series test

If $(a _ k)$ is a monotonically decreasing sequence and $a _ k \to 0$, then $\sum^\infty _ {k=1}(-1)^{k+1}a _ k$ converges.


Regroup terms of $s _ {2n}$ and $s _ {2n+1}$ to show they’re bounded monotonic subsequences.

When proving the alternating series test,

If $(a _ k)$ is a monotonically decreasing sequence and $a _ k \to 0$, then $\sum^\infty _ {k=1}(-1)^{k+1}a _ k$ converges.

You work with the partial sums $s _ {2n}$ and $s _ {2n+1}$. What two ways do you rearrange

\[\begin{aligned} s_{2n} &= a_1 - a_2 + a_3 - a_4 + \ldots - a_{2n-2} + a_{2n-1} - a_{2n} \end{aligned}\]

to show it’s both monotonically increasing and bounded above?


\[\begin{aligned} s_{2n} &= a_1 - a_2 + a_3 - a_4 + \ldots - a_{2n-2} + a_{2n-1} - a_{2n} \\\\ &= (a_1 - a_2) + (a_3 - a_4) + \ldots + (a_{2n-1} - a_{2n}) \\\\ &= a_1 - (a_2 - a_3) - \ldots - (a_{2n-2} - a_{2n-1}) - a_{2n} \\\\ \end{aligned}\]

When proving the alternating series test,

If $(a _ k)$ is a monotonically decreasing sequence and $a _ k \to 0$, then $\sum^\infty _ {k=1}(-1)^{k+1}a _ k$ converges.

You work with the partial sums $s _ {2n}$ and $s _ {2n+1}$. What two ways do you rearrange

\[\begin{aligned} s_{2n+1} &= a_1 - a_2 + a_3 - a_4 + \ldots + a_{2n-1} - a_{2n} + a_{2n+1} \end{aligned}\]

to show it’s both monotonically decreasing and bounded below?


\[\begin{aligned} s_{2n+1} &= a_1 - a_2 + a_3 - a_4 + \ldots + a_{2n-1} - a_{2n} + a_{2n+1} \\\\ &= a_1 - (a_2 - a_3) - \ldots - (a_{2n} - a_{2n+1}) \\\\ &= (a_1 - a_2) + (a_3 - a_4) + \ldots + (a_{2n-1} - a_{2n}) + a_{2n+1} \end{aligned}\]

When proving the alternating series test,

If $(a _ k)$ is a monotonically decreasing sequence and $a _ k \to 0$, then $\sum^\infty _ {k=1}(-1)^{k+1}a _ k$ converges.

You end up showing that $s _ {2n} \to L$ and $s _ {2n+1} \to U$ both have limits. How do you show they’re equal?


By the AOL,

\[\begin{aligned} U - L &= \lim_{n \to \infty} s_{2n+1} - s_{2n} \\\\ &= \lim_{n\to\infty} a_{2n+1} \\\\ &= 0 \end{aligned}\]



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