amyna
  • amyna
Summation from 0 to infinity K+1/3^k Do I use the ratio test? But how do I do it?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jango_IN_DTOWN
  • jango_IN_DTOWN
what is the question?
jango_IN_DTOWN
  • jango_IN_DTOWN
@amyna
amyna
  • amyna
It's K+1/3^k

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jango_IN_DTOWN
  • jango_IN_DTOWN
convergence/ divergence test?
jango_IN_DTOWN
  • jango_IN_DTOWN
\[k+\frac{ 1 }{ 3^k }\]
jango_IN_DTOWN
  • jango_IN_DTOWN
this one??
amyna
  • amyna
It says if this converges absoulty, conditionally, or just diverged
amyna
  • amyna
No k+1 is in the numerator
jango_IN_DTOWN
  • jango_IN_DTOWN
\[\sum_{0}^{\infty} \frac{ k+1 }{ 3^k }\]
amyna
  • amyna
Yes
jango_IN_DTOWN
  • jango_IN_DTOWN
do the ratio test first...
jango_IN_DTOWN
  • jango_IN_DTOWN
t_k= (k+1)/3^k and t_k+1= (k+2)/3^(k+1)
amyna
  • amyna
Oh okay got it! I was just confused if I had to add the k+1 in the numerator or not
amyna
  • amyna
Thanks!!
jango_IN_DTOWN
  • jango_IN_DTOWN
replace k by k+1 to get t_k.. so whats the solution? do you need more help?
jango_IN_DTOWN
  • jango_IN_DTOWN
its convergent :)
amyna
  • amyna
The numerator becomes k+2 right?
jango_IN_DTOWN
  • jango_IN_DTOWN
which form do you use ? \[\lim |\frac{ a_n }{ a_{n+1} }| \]
jango_IN_DTOWN
  • jango_IN_DTOWN
or \[\lim |\frac{ a _{n+1} }{ a_n }|\] ?
jango_IN_DTOWN
  • jango_IN_DTOWN
@amyna
amyna
  • amyna
The second one
jango_IN_DTOWN
  • jango_IN_DTOWN
ok \[\lim |\frac{ t _{k+1} }{ t_k }|=\lim |\frac{ (k+2). 3^k }{ (k+1) .3^{k+1}}|\]
jango_IN_DTOWN
  • jango_IN_DTOWN
= \[\lim |\frac{ 1 }{ 3 }. \frac{ 1+1/k }{ 1+2/k }|=1/3\]
jango_IN_DTOWN
  • jango_IN_DTOWN
<1 so by ratio test it is ?????
jango_IN_DTOWN
  • jango_IN_DTOWN
@amyna
jango_IN_DTOWN
  • jango_IN_DTOWN
in the numerator it will be 1+2/k and in the denominator 1+1/k

Looking for something else?

Not the answer you are looking for? Search for more explanations.