Question

Use Alternative series to determine convergence

Answer 1

\[\sum_{n=1}^{\infty}\frac{n}{n^3+16}\]

Answer 2

I forgot (-1)^n

Answer 3

Probably.

Answer 4

probably is not an answer i'd wan to give my teacher on my final exam

Answer 5

it converges absolutely

Answer 6

the limit to 0 condition can be solved with lhopitals

Answer 7

but i'm wondering about An+1 < An

Answer 8

waste to use alternating series test

Answer 9

for An+1<An

try letting \[f(n)=\frac{n}{n^3+16}\]

show that f'(n)<0 (for n>3)

Answer 10

waste what do you mean? and The test says "USE ALTERNATIVE" not whatever the hell i want

Answer 11

it is a waste for this problem since it is absolutely convergent trivially by the comparison test

Answer 12

if you do what i posted above you can show that it is at least conditionally convergent using the AST

Answer 13

your teacher/prof/book is being lazy in picking a crappy problem to use the AST on.

Answer 14

and you can prove that this is absolutely convergent

Answer 15

by limit comparison

Answer 16

not conditionally

Answer 17

yes...very easy

Answer 18

it's just the An+1<An

Answer 19

you need that for the AST.

Answer 20

not for comparison/integral/ration/root...tests

Answer 21

wouldn't An+1 be\[\frac{n+1}{(n+1)^3+16}\]

Answer 22

yes

Answer 23

well all values of n have to be less than an right?

Answer 24

no

Answer 25

that's what my book says >_<

Answer 26

for the AST...there needs to be an N such that for all n>N

\[A_{n+1}<A_n\]

Answer 27

your book in wrong

Answer 28

is it? that would explain this is what my book says

Answer 29

do you have a book to say otherwise?

Answer 30

cause my book once again says that to Apply The alternating series for n>= 1

Answer 31

that is true ... but what I wrote is a stronger condition that is also true

Answer 32

think about this...

Answer 33

well how can it be true if n can be >= 1 ... you put one and the statement is false... but any larger intger works

Answer 34

suppose that you have a sequence \[a_n\]
with \[a_n>0\] for all n

that is increasing for n=1...10 and decreasing to zero afterwards...

Answer 35

alright?

Answer 36

and it does say that you can modify for all n greater than some integer N

Answer 37

then \[\sum_{n=1}^{\infty}(-1)^na_n\]

\[=\sum_{n=1}^{10}(-1)^na_n+\sum_{n=11}^{\infty}(-1)^na_n\]

Answer 38

teacher just never explained it.. and it's a note at the end of the book

Answer 39

\[\sum_{n=1}^{10}(-1)^na_n\]

is a finite sum so convergent

and

\[\sum_{n=11}^{\infty}(-1)^na_n\]

converges by the AST

Answer 40

for your sequence \[a_n=\frac{n}{n^3+16}\]

starts decreasing after 2

\[a_1>a_2\]

but that doesn't matter

Answer 41

so in order to modify is there any steps inbetween doing so or does it just hold up that way

Answer 42

i mean \[a_1<a_2\]

Answer 43

like for teacher purposes shoul di say something like An+1<=An for all N>=2

Answer 44

then \[a_n+1<a_n\] for all \[n\geq 2\]

Answer 45

yes

Answer 46

all you care about is the tail of the sequence

Answer 47

alright thanks

Answer 48

np