Thread: Convergence of Sequences and Series - My Brain just doesnt get it

1. Convergence of Sequences and Series - My Brain just doesnt get it

Hello folks. First off i want to say this is an awesome forum! But lets get to the topic:

I totally understand the concept of convergence/divergence and i dont have problems to follow the lessons. But....
when it comes down to exercises/exams i totally fail in showing wether a series/sequence converges or not

for example:

i got the series: $\sum_{n=1}^{\inf}\frac{cos(4n)}{n}$

my thinking was... finding a series a_n greater than the given one and then proof its convergence. but i seem to fail ....

another typical problem would be:

$lim(1+\frac{2}{n})^{4n}$ testing for convergence as n goes to infinity

my thought was writing the 1 as $\frac{n}{n}$ and then have $\frac{2+n}{n}$ raised to the power of 4n

but to be honest... im lost !

2. $\sum_{n=1}^{\infty}\frac{cos(4n)}{n} \leq \sum_{n=1}^{\infty}\frac{1}{n}$

since -1 <= cosf(x) <= 1

and the second series diverge by integral test

3. Originally Posted by Amer
$\sum_{n=1}^{\infty}\frac{cos(4n)}{n} \leq \sum_{n=1}^{\infty}\frac{1}{n}$

since -1 <= cosf(x) <= 1

and the second series diverge by integral test
But if something is less than a divergent series, it may as well be convergent or divergent.
It is not possible to conclude.

Moreover, this kind of series comparison only works for series with a positive general term.
And since $\cos(4n)$ can be positive or negative, this method doesn't work here.

So we may use some things dealing with alternating series.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$a_n$, $b_n$ positive sequences.

If $\sum a_n\leq \sum b_n$, then
- if $\sum b_n$ converges, $\sum a_n$ converges
- if $\sum a_n$ diverges, $\sum b_n$ diverges

4. if the series is alternating then it is converge according to this theorem
alternating series is converges if the following two condition are stratified
$1) a_1>a_2>a_3>.....>a_k...
$

$
2) lim_{k\rightarrow+\infty}a_k=0
$

in this series a1>a2>a3
and the limit of

$lim_{k\rightarrow+\infty} \frac{cos4k}{k}=0
$

so the series is converge

5. note that the series, $\sum_{n=1}^{\infty} \frac{\cos(4n)}{n}$ , is not strictly alternating.

6. Originally Posted by coobe
for example:

i got the series: $\sum_{n=1}^{\inf}\frac{cos(4n)}{n}$
This is not the simplest example by far! It is not exactly alternating, just "kind of".

The tool for this one is the Abel transformation, which allows to prove the following theorem:

If $(u_n)_n$ decreases to 0, and $(a_n)_{n\geq 1}$ is a sequence of complex numbers such that the partial sums $\left|\sum_{n=1}^N a_n\right|$ are bounded (i.e. $\exists M>0, \forall N\geq 1, \left|\sum_{n=1}^N a_n\right|\leq M$), then the series $\sum_n a_n u_n$ converges.

In your case, $u_n=\frac{1}{n}$ and $a_n=\cos(4n)$. In order to prove that $a_n$ satisfies the hypothesis, use the fact that $a_n=\Re(e^{4in})$ and note that $\sum_{n=1}^N e^{4in}$ is a sum of terms in a geometric sequence.

7. well a huge problem might be that the exercise says: "Use the comparison test" (Majoranten/minorantenkriterium in german)

8. Originally Posted by coobe
well a huge problem might be that the exercise says: "Use the comparison test" (Majoranten/minorantenkriterium in german)
Are you sure it is not $\sum_n \frac{\cos(4n)}{n^2}$ for instance? Or did you consider similar series right before, like $\sum_n\frac{\cos n}{n}$, that could be re-used? (This wouldn't help performing a comparison : here, this is impossible. But other tricks may come in handy)

9. no it is simply n without power. it is actually an exercise from an old caculus1 exam. we use it to prepare for future ones so i dont think youll need to know a similar series

10. Three steps...

$\sum_{n=1}^{\infty} \frac{x^{n}}{n} = \ln (1-x)$ (1)

$\cos 4n = \frac{e^{4in}+e^{-4in}}{2}$ (2)

$\sum_{n=1}^{\infty} \frac{\cos 4n}{n}= \sum_{n=1}^{\infty}\frac{e^{4in}+e^{-4in}}{n}= \frac{1}{2}\cdot \{\ln (1-e^{4i}) + \ln (1-e^{-4i})\}$ (3)

Kind regards

$\chi$ $\sigma$