Question

series of (-1)* 3n^2+1/5n+2

Answer 1

\[ \sum_{1}^{\infty}(-1)^n/\sqrt[3]{n}
is it converges or diverges\]

Answer 2

@Zarkon someone needs u :)

Answer 3

Is this your series?\[\sum_{n=0}^{\infty} (-1)^n \frac {3n^2 + 1}{5n + 2}\]

Answer 4

yes

Answer 5

is it diverges or converges

Answer 6

It diverges

Answer 7

by which test

Answer 8

Dunno the name of the test, but since its an alternating series in the form of \[\sum_{n=0}^{\infty} (-1)^n a_n\]First thing you have to show for it to converge is that \[\lim_{n \rightarrow \infty} a_n = 0\]But for this, the limit is infinity, so the thing diverges just from that.

Answer 9

i got it thank am from reno where r u from

Answer 10

np, I'm from nyc =)

Answer 11

nyc ftw =)

Answer 12

\[\sum_{1}^{\infty}(-1)^n/\sqrt[3]{n}\]

Answer 13

converges

Answer 14

which test

Answer 15

its a harmonic with p = 1.5. Since p >1, it converges, the alternation makes it converge even faster.

Answer 16

it is just called the divergence test.

Answer 17

oh i see

Answer 18

the second one is the alterneting series test.

Answer 19

is it conditionally

Answer 20

no, its absolute. There are no x terms for it to be conditional convergence.

Answer 21

this

\[\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}\] diverges

but

\[\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}\] converges by the AST

Answer 22

:o the first diverges? didn't know.

Answer 23

yes...p-series test.

Answer 24

but p = 1.5, so...

Answer 25

p=1/3

Answer 26

so its only p-series and ABS conv noALT

Answer 27

I thought it was \[\sum_{n = 0}^{\infty} \frac {1}{n^P}\]

Answer 28

\[\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}=\sum_{n=1}^{\infty}\frac{1}{n^{1/3}}\]

Answer 29

oh, right, what am I thinking :3

Answer 30

you cant start at zero either

Answer 31

SO I THINK IS CONV p-series and ALT series right

Answer 32

thanks, my mind is so wacky today ;/

alternating series test.

Answer 33

the series is conditionally convergent

Answer 34

why is conditionally cqan u tel me so i can understanfd

Answer 35

\[\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}\] converges by the AST

\[\sum_{n=1}^{\infty}\left|\frac{(-1)^n}{\sqrt[3]{n}}\right|=\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}\] diverges by the p-series test.

therefore it is conditionally convergent

Answer 36

\[x \cos (x) what is the series\]

Answer 37

ok

Answer 38

is that \[\sum_{0}^{\infty}(-1)^n/(2n_1)!*x^2n+1\]

Answer 39

is that right

Answer 40

I think I know what you wrote..latex looks bad...I believe it is ok

Answer 41

oh that was 2n+1

Answer 42

can u corct it plz zarkon

Answer 43

\[\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n+1}\]

Answer 44

this look like e^x series

Answer 45

but anm try to wirte xcosx series

Answer 46

is there anyone can help

Answer 47

r u there zarkon

Answer 48

The Maclauren series for cosine is\[\cos x = 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!} + ...\]So when you multiply cosine x by x, your series is \[x \cos x = x \sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac {x^{2n+1}}{(2n)!}\]

Answer 49

thanks