Question

d

Answer 1

are u sure 1st term is 8 and not 6??

Answer 2

Yeah

Answer 3

let us split this into 2,like this
5+(3+2+4/3+8/9+......)
i will find the sum of the geometric series in bracket....and then add 5..
ok with this?

Answer 4

are you allowed to do that?

Answer 5

yes,i am,because it does not change the total sum....

Answer 6

okay so how would you get the series for the inside

Answer 7

1st u realize that it is geometric sequence with common ratio 2/3??

Answer 8

okay i see that now

Answer 9

the top seems like it is moving faster

Answer 10

so the formula for sum of infinite series is a/(1-r)
where a is 1st term and r is common ratio....
so can u find the sum now?

Answer 11

so it would be 14?

Answer 12

i dont get how you know you can separate 5 +3 how come you dont do 6+2 or something like that

Answer 13

lets see a=3,r=2/3
\[\frac{ a }{ 1-r }=\frac{ 3 }{ 1-\frac{ 2 }{ 3 } }=\frac{ 3*3 }{ 3-2 }=9\]

Answer 14

that is because, the common ratio was 2/3....so the term before 2 was 2*3/2=3
that is why i wanted 3,so i split 8 as 5+3....

Answer 15

alright i think i got it. thanks

Answer 16

so u got what it converges to??

Answer 17

14

Answer 18

right?

Answer 19

yes :) 9+5=14...correct

Answer 20

thank you. should i ask another question about series here or a different thread?

Answer 21

as u wish :)
preferably different thread,so that everyone can help
but i don't mind if u put it here

Answer 22

ill just ask here and if you cant ill ask everyone :)

Answer 23

\[\sum_{k=1}^{infiniti} ((-1)^k \ln(k) )/ \ln(k+k^2)\]

Answer 24

does the converge if so what type or diverge or do we not have enough info for it?

Answer 25

nopes,sorry...weak in summation.....ask everyone....

Answer 26

alrighty.