Question

Summation from 0 to infinity

K+1/3^k

Do I use the ratio test? But how do I do it?

Answer 1

what is the question?

Answer 2

@amyna

Answer 3

It's
K+1/3^k

Answer 4

convergence/ divergence test?

Answer 5

\[k+\frac{ 1 }{ 3^k }\]

Answer 6

this one??

Answer 7

It says if this converges absoulty, conditionally, or just diverged

Answer 8

No k+1 is in the numerator

Answer 9

\[\sum_{0}^{\infty} \frac{ k+1 }{ 3^k }\]

Answer 10

Yes

Answer 11

do the ratio test first...

Answer 12

t_k= (k+1)/3^k and t_k+1= (k+2)/3^(k+1)

Answer 13

Oh okay got it! I was just confused if I had to add the k+1 in the numerator or not

Answer 14

Thanks!!

Answer 15

replace k by k+1 to get t_k.. so whats the solution? do you need more help?

Answer 16

its convergent :)

Answer 17

The numerator becomes k+2 right?

Answer 18

which form do you use ?
\[\lim |\frac{ a_n }{ a_{n+1} }| \]

Answer 19

or
\[\lim |\frac{ a _{n+1} }{ a_n }|\] ?

Answer 20

@amyna

Answer 21

The second one

Answer 22

ok
\[\lim |\frac{ t _{k+1} }{ t_k }|=\lim |\frac{ (k+2). 3^k }{ (k+1) .3^{k+1}}|\]

Answer 23

=
\[\lim |\frac{ 1 }{ 3 }. \frac{ 1+1/k }{ 1+2/k }|=1/3\]

Answer 24

<1 so by ratio test it is ?????

Answer 25

@amyna

Answer 26

in the numerator it will be 1+2/k and in the denominator 1+1/k