Question

converges or diverges
summation symbol from 1 to infinity (0.001 + (k^(-1.5)
Use divergence test

Answer 1

\[\sum_{k=1}^{\infty}0.001+k^{-3/2}\]?

Answer 2

\[\sum_{1}^{\infty} 0.001 + k ^{-1.5}\]

Answer 3

yeah i'm not sure how to use equation editor when posting a question.
i'm still new to this

Answer 4

no prob, better to learn the LaTeX language when you get the chance
what kind of test are you supposed to use here? there are a few "divergence tests" as far as I know

Answer 5

is this the one where we take the limit of the partial sums?
hm.... I'm not the best at these
@satellite73 @amistre64 series problem

Answer 6

i know we are suppose to take the limit of \[a_{k}\] and see if it equals zero, then we know its convergent.
but thats all i know

Answer 7

wouldn't your \[a_{k}\] = \[k^{-1.5}\]

Answer 8

ok if,
\[\lim_{k \rightarrow \infty}.001+k^{-1.5}=0 \]
then it will be convergent..
so let's do that:
\[\lim_{k \rightarrow \infty}.001+k^{-1.5}=\lim_{k \rightarrow \infty}(1/1000)+(1/k^{1.5})\]
\[\lim_{k \rightarrow \infty} (k^{1.5}+1000)/(1000k^{1.5})\]
to make things easier I'll use the L'Hopital stuff...
\[=\lim_{k \rightarrow \infty} (1.5k^{.5})/(1.5)(1000k^{.5})=\lim_{k \rightarrow \infty} 1/1000=1/100\]
since
\[\lim_{k \rightarrow \infty} .001+k^{-1.5} \neq 0\]
then the series will diverge

Answer 9

@mrlunarugby the a sub n will be .001+k^(-1.5)

Answer 10

so suppose if we have
\[\sum_{n=1}^{\infty}(oaiwejnofivjawelravefav)^n+n^2+e^n+logn\]
our a sub n will be

oaiwejnofivjawelravefav)^n+n^2+e^n+logn
and to find if this series diverges, we find the
\[\lim_{n \rightarrow \infty} a_n=\lim_{n \rightarrow \infty} oaiwejnofivjawelravefav)^n+n^2+e^n+logn\]
if
\[\lim_{n \rightarrow \infty} oaiwejnofivjawelravefav)^n+n^2+e^n+logn=0\]
then it is convergent. otherwise, it will be divergent

Answer 11

Right. thanks it took me a minute to understand it.
the way you changed 0.001 to 1/1000 then k^(-1.5) to 1/k^1.5
i just didn't think to do it that way. that was very helpful. thanks
and by the way in the question that i asked you shouldn't the limit approach 1/1000 not 1/100.
was that a typo??

Answer 12

how do i type equations when i'm posting questions?
because i have more problems i need help with.

Answer 13

gotta learn the language LaTeX to type equations quickly

Answer 14

ohh is that a programming language

Answer 15

i'm only learning java at the moment

Answer 16

it's not that hard, here's a cheat sheet:
http://omega.albany.edu:8008/Symbols.html

Answer 17

oh thanks. i can't learn all this know though i'm trying to cram for final

Answer 18

now*

Answer 19

you have to enclose things in brackets
i.e.
\sum_{i=1}^{\infty}i
enclose this in brackets like so
\[\sum_{i=1}^{\infty}i\.]
^without this dot
and it becomes...

Answer 20

\[\sum_{i=1}^{\infty}i\]

Answer 21

but anyway, anonymous found that the series diverges because the limit of the summand as \(k\to\infty\neq0\)
that is how that trick worked

Answer 22

yeah that was very helpful

Answer 23

mind you that if\[\lim_{k\to\infty}\{a_n\}=0\]that does \9not\) mean that the series does converge, only that it might converge

Answer 24

*does not

Answer 25

for example\[\sum_{i=1}^{\infty}\frac1i\]diverges\[\sum_{i=1}^{\infty}\frac1{i^2}\]converges

Answer 26

@mylunarugby oh yeah :)) my bad, either it was a typo or there was not enough space :))