Alternating series test question.

Question

I'm reading this in my text:

So the alternating series test says:

i) is about the sequence decreasing

ii) is about the limit of the b term going to 0

I'm confused about why we need to do anything more once we find out that ii) isn't satisfied in example 2. What does it mean that we're looking at the limit of the nth term of the series? Don't we know that it diverges already? Also, can someone show me how they determine the limit of $a_n$?

Answer 1

Answer to First Question: So, notice that the summand in example 2 has the form $$(-1)^n b_n = (-1)^n \frac{3n}{4n-1}$$ where, clearly, $b_n = \frac{3n}{4n-1}$. This sequence $\{b_n\} = \{\frac{3n}{4n-1}\}$ is the one we must consider in the second condition for the alternating series test.

One condition that we have to check in order to use the alternating series test is that $$\lim b_n = 0$$ However, as the author pointed out, $$\lim b_n = \lim \frac{3n}{4n-1} = \frac{3}{4} \neq 0$$ Therefore, we cannot use the alternating series test to conclude anything; we have to try something else.

Answer to Question 2: To consider the $n$th term of the series means taking an arbitrary term in the series which has the form $(-1)^n \frac{3n}{4n-1}$. Notice that this includes the $(-1)^n$, as opposed to just the $\frac{3n}{4n-1}$. We then take $$\lim (-1)^n \frac{3n}{4n-1}$$

Answer to Question 3: It should be clear this sequence $\big\{(-1)^n \frac{3n}{4n-1}\big\}$ diverges due to oscillation, but we can be a bit more explicit. If the sequence converged, then if we consider the subsequences when $n$ is odd and $n$ is even, they should converge to the same limit.

However, if $n$ is even, then $$n \text{ even} \implies \lim (-1)^n \frac{3n}{4n-1} = \lim \frac{3n}{4n-1} = \frac{3}{4}$$ and if $n$ is odd, $$n \text{ odd} \implies \lim (-1)^n \frac{3n}{4n-1} = \lim - \frac{3n}{4n-1} = - \frac{3}{4}$$ Therefore, since these two subsequences do not converge to the same value, it must be that the sequence $\big\{(-1)^n\frac{3n}{4n-1}\big\}$ diverges. Hence, by the divergence test, the series in question diverges.

Answer 2

Your confusion is justified.

From the definition of alternating series $\sum_{n=1}^\infty a_n$: $$ a_n=(-1)^nb_n,\quad b_n>0, $$ it follows that $|a_n|=b_n$. Therefore $$ \lim_{n\to\infty} b_n=0 \iff \lim_{n\to\infty} |a_n|=0.\tag1 $$ If the condition $(1)$ is not satisfied $\sum a_n$ does not converge as the condition is necessary for convergence of any series.

Therefore if alternating series test fails by this reason, no further tests are needed. The series does not converge.

Answer 3

If conditions (i) and (ii) are satisfied, then you conclude that the series ${\bf converges}$.

If one of the conditions fails, then you cannot conclude that the series ${\bf diverges}$

The limit $(a_n)$ does not exist because it converges to two different values. In fact, if $n$ is even, then sequence $a_n$ converges to $3/4$ while it converges to $-3/4$ for the odd terms and thus it diverges. In particular, it $\lim a_n \neq 0$ and so series diverges by divergence test.

Answer 4

Here is a discussion of a more general case of this.

Suppose $b_n \ge b_{n+1}$ and $\lim_{n \to \infty} b_n =b > 0 $.

Let $s_n =\sum_{k=1}^n (-1)^nb_n $.

How does $s_n$ behave?

Let $t_n =\sum_{k=1}^n (-1)^n(b_n-b) $. Since $b_n-b \to 0$ and $b_n-b$ is decreasing, $\lim_{n \to \infty} t_n$ exists. Call it $T$.

$t_n =\sum_{k=1}^n (-1)^kb_k+\sum_{k=1}^n (-1)^kb =\sum_{k=1}^n (-1)^kb_k+b\sum_{k=1}^n (-1)^k $ so $\sum_{k=1}^n (-1)^kb_k =t_n-b\sum_{k=1}^n (-1)^k $.

Therefore

$\begin{array}\\ \sum_{k=1}^{2n} (-1)^kb_k &=t_{2n}-b\sum_{k=1}^{2n} (-1)^k\\ &=t_{2n}\\ &\to T\\ \text{and}\\ \sum_{k=1}^{2n+1} (-1)^kb_k &=t_{2n+1}-b\sum_{k=1}^{2n+1} (-1)^k\\ &=t_{2n+1}+b\\ &\to T+b\\ \end{array} $

Note that $\sum_{k=1}^{2n} (-1)^kb_k =\sum_{k=1}^{n} (b_{2k}-b_{2k-1}) \to T $.

Your case is $b_n =\dfrac{3n}{4n-1} $ so, as you wrote $b = \dfrac34 $ so the limit points are $T$ and $T+\dfrac34$ where

$\begin{array}\\ T &=\lim_{n \to \infty} \sum_{k=1}^{n} (b_{2k}-b_{2k-1})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{6k}{8k-1}-\dfrac{3(2k-1)}{4(2k-1)-1})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{6k}{8k-1}-\dfrac{6k-3}{8k-5})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{6k(8k-5)-(6k-3)(8k-1)}{(8k-1)(8k-5)}\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{48k^2-30k-(48k^2-30k+3)}{(8k-1)(8k-5)}\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{-3}{(8k-1)(8k-5)}\\ &=\lim_{n \to \infty} -\dfrac{3}{64}\sum_{k=1}^{n}\dfrac{1}{(k-1/8)(k-5/8)}\\ \end{array} $

Note that if $b_n =\dfrac{un}{vn+w} $ then $b = \dfrac{u}{v} $ so the limit points are $T$ and $T+\dfrac{u}{v}$ where

$\begin{array}\\ T &=\lim_{n \to \infty} \sum_{k=1}^{n} (b_{2k}-b_{2k-1})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{u(2k)}{v(2k)+w}-\dfrac{u(2k-1)}{v(2k-1)+w})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{2uk}{2vk+w}-\dfrac{2uk-u}{2vk-v+w})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{2uk(2vk-v+w)-(2uk-u)(2vk+w)}{(2vk+w)(2vk-v+w)}\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{4uvk^2+2u(w-v)k-(4uvk^2+2k(uw-uv)-uw)}{(2vk+w)(2vk-v+w)}\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{uw}{(2vk+w)(2vk-v+w)}\\ &=\lim_{n \to \infty} \dfrac{uw}{4v^2}\sum_{k=1}^{n}\dfrac{1}{(k+w/(2v))(k+(w-v)/(2v))}\\ &=\lim_{n \to \infty} \dfrac{uw}{4v^2}\sum_{k=1}^{n}\dfrac{1}{(k+w/(2v))(k+w/(2v)-1/2)}\\ \end{array} $