Question

please help! does it converge or diverge??? summation from 0 to infinity of [(4)/(ln(n+5))*(cos((npi)/2))]

Answer 1

lol what a gimmick

Answer 2

is it clear that \[\cos(\frac{n\pi}{2})\] is just another way of writing an alternating series?

Answer 3

oh wait i am wrong!! it doesn't alternate damn

Answer 4

it is 0,-1,0,1,...

Answer 5

pretty sure it does NOT converge,
not sure why

Answer 6

maybe we can say that it has two sub sequnces that do not converge

Answer 7

think that does it right?

Answer 8

It looks like an alternating series with log on the bottom.
|terms| decreases, and thus it would converge

Answer 9

That is my first look at it

Answer 10

\(\large\color{black}{\displaystyle 4\sum_{i=1}^{\infty }\frac{\cos(i \pi/2)}{\ln(n+5)}}\)

it would alternate, but after every term (skipping one term because it zero)

+ , 0, - , 0, +, 0, -, etc...

Answer 11

but it converges

Answer 12

Lets observe the following series:

\(\large\color{black}{\displaystyle \sum_{k=0}^{\infty }\frac{\cos(k \pi/2)}{\ln(k+5)}}\)


This is the behavior of each component of the series:

\(\large{\bbox[5pt,#CCFFFF ,border:2px solid black ]{ \displaystyle \left.\begin{matrix} k & & \cos\left(\dfrac{k\times \pi}{2}\right) & &\ln\left(k+5\right) \\
\\[0.5em]
k=0 & | & 1&| & \ln(5) \\[0.5em]
k=1 & | & 0&| & \ln(6)\\[0.5em]
k=2 & | & -1&| & \ln(7)\\[0.5em]
k=3 & | & 0&| & \ln(8)\\[0.5em]
k=4 & | & 1&| & \ln(9)\\[0.5em]
k=5 & | & 0&| & \ln(10)\\[0.5em]
k=6 & | & -1&| & \ln(11)\\[0.5em]
k=7 & | & 0&| & \ln(12)\\[0.5em]
k=8 & | & 1&| & \ln(13)\\[0.5em]
{\rm ~...} && {\rm ~...}\end{matrix}\right. }}\)

So we can write the terms out:
(I will omit the zeros.)

\(\color{black}{ \displaystyle \frac{ 1}{\ln[5]}-\frac{ 1}{\ln[7]}+\frac{ 1}{\ln[9]}-\frac{ 1}{\ln[11]}+\frac{ 1}{\ln[13]} -\frac{ 1}{\ln[15]}{\bf ...}}\)


\(\large{\bbox[5pt, #fff999 ,border:2px solid black ]{ \rm Alternating~Series~Test: }}\)
By the alternating series test the series converges:
That is:

\(\color{black}{\bf \left.Condition\text{#}1\right) }\) \(\color{black}{ \left|a_{k}\right| >\left|a_{k+1}\right|}\)

\(\color{black}{\bf \left.Condition\text{#}2\right) }\) \(\color{black}{ \displaystyle \lim_{k\to\infty}\left|a_{k}\right| =0}\)

Also, and obviously: \(\color{black}{\bf \left.Condition\text{#}3\right) }\)
The terms alternate, and none
of the terms are undefined.
\(\scriptsize\color{ slate }{\scriptsize{\bbox[5pt, #fff999 ,border:2px solid royalblue ]{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ }}}\)


`And our series in fact satisfies these conditions.`

`So, THIS SERIES CONVERGES.`

Also, if you want to write an equivalent
but more convenient representation for
this series somehow, you can write this
series as:

\(\large\color{black}{\displaystyle \sum_{k=0}^{\infty }\frac{(-1)^k}{\ln(2k+5)}}\)