# Valid conclusion for an absolutely convergent sequence

• cherry
cherry
Homework Statement
Determine whether the series is absolutely convergent,
conditionally convergent, or divergent. (see description)
Relevant Equations
See description

Hello, this is my attempt for #19 for 11.6 of Stewart's “Multivariable Calculus”.
The question is to determine whether the series is absolutely convergent, conditionally convergent, or divergent.
The answer solutions used a ratio test to reach the same conclusion but I used the comparison test.
Is my method also valid? Thanks.

That is not valid. You say that ##\cos( \frac{\pi}{3} n) \ge n/2##. That does not stop it from being huge and the series being divergent. You should be showing that its absolute value is less than something. You should be able to prove a simple upper limit for its absolute value.

Last edited:
cherry and docnet
It seems like you were trying to use a comparison test, but accidentally reversed the inequality sign. Aside from that error, there's another flaw.
Given a series $$S=\sum_{n=1}^\infty a_n,$$
$$\lim_{n\to \infty } a_n=0 \text{ doesn't mean } S \text{ converges}.$$
For example,
$$\lim_{n\to \infty} \frac{1}{n}=0 \text{ and } \sum_{n=1}^\infty \frac{1}{n} \text{ is divergent}.$$
Aside from deriving an upper limit of the series, it's also pretty straightforward to use the ratio test.

cherry
Your reasoning is strange. On top of what's already mentioned, you write something like "convergent, therefore absolutely convergent". This is not true.

For sufficiently large indices it holds that
$$\left\lvert \frac{\cos f(n)}{n!}\right\rvert \leqslant \frac{1}{n^2}.$$
What do we conclude?

docnet
To further explain post #4, since it looks like the OP needs more help:

A convergent series is not necessarily absolutely convergent, and a function that's absolutely convergent is convergent.

With this approach you need the fact that the following sum converges.

$$\sum_{n=1}^\infty \frac{1}{n^2}$$

See the Basel problem on wiki.

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