Question

determine whether the series converges absolutely, converges conditionally or diverges

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^n *(2n)!}{ 5n! }\]

Answer 2

i got an = (2n)!/5n! i dont know what the ! symbols mean

Answer 3

means "factorial"

Answer 4

what do i do with it?

Answer 5

like how do i determine if converges absolutely or conditionally?

Answer 6

nothing really \((2n)!\) is much larger than \(5n!\) the terms do not even go to zero

Answer 7

this is a divergent series
take the limit

Answer 8

so if they dont go to zero then that means that it diverges?

Answer 9

@satellite73 like this \[\lim_{n \rightarrow \infty} \frac{ (2n)! }{ 5n! }\]

Answer 10

@rational could help me with this?

Answer 11

what do i with the factorial symbol?

Answer 12

are you sure the denominator is 5*n! and not (5n)! ?

Answer 13

yeah its 5n!

Answer 14

take a screenshot and attach maybe..

Answer 15

okay

Answer 16

here it is @rational

Answer 17

yes it does not converge

you may use ratio test to conclude that

Answer 18

we havent covered ratio test yet :(

Answer 19

for this other one \[\frac{ 16^\frac{ 1}{ 2 } }{ 7^{n+1}}\]

Answer 20

what do i do with the n+1 exponent?

Answer 21

you want to know whether it converges or diverges ?

Answer 22

here is the whole problem

Answer 23

no whether it coverges absolutely conditionally or diverges?

Answer 24

simplify this first \[\frac{16^{n/2}}{7^{n+1}}\]

Answer 25

\[\frac{16^{n/2}}{7^{n+1}} = \frac{(4^2)^{n/2}}{7\cdot 7^{n}} = \dfrac{1}{7}\cdot \frac{4^n}{7^n} = \frac{1}{7}\cdot \left(\frac{4}{7}\right)^n\]

Answer 26

does that look familair ?

Answer 27

sort of looks like a harmonic series right?

Answer 28

\[\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{16^{n/2}}{7^{n+1}} ~~=~~\frac{1}{7}\sum\limits_{n=1}^{\infty} (-1)^{n-1}\left(\frac{4}{7}\right)^{n} \]

Answer 29

it is a geometric series with a common ratio of \(\frac{4}{7}\)

Answer 30

does the geometric series with common ratio 4/7 converges or diverges ?

Answer 31

converges right?

Answer 32

Yep! and it converges absolutely

Answer 33

how do you determine its absolutely?

Answer 34

because the absolute value of the series :
\[\left|\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{16^{n/2}}{7^{n+1}} \right|\]

is same as geometric series which converges

Answer 35

if the absolute value of a series converges, then the original series also converges. then we say that the "series converges absolutely"

Answer 36

and to see if the absolute value of the series converges we find its limit right?

Answer 37

what we found above is that absolute value of the given series converges

Answer 38

\[\left|\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{16^{n/2}}{7^{n+1}} \right|~~~ =~~~\sum\limits_{n=1}^{\infty}\frac{16^{n/2}}{7^{n+1}} \]

Answer 39

that right side is a converging geometric series because the ratio is 4/7

that proves the absolute value of given series converges

Answer 40

oh got it

Answer 41

for the last problem with the factorial symbols how did we proof it diverged again?

Answer 42

thanks for your help @rational