Question

(1/5)+(1/8)+(1/11)+(1/14)+(1/17)...+...
Determine whether the series is convergent or divergent. This problem is in the integral test part of my book. I notice that the pattern is consistently going up by 3, but cant seem to figure out (asubn).

Answer 1

if i know (asubn) then i can convert it to an integral and be able to solve it from there, but am having trouble figuring out the pattern...

Answer 2

Took me a second, just wanted to make sure I could get the right answer for ya (bit rusty). So your series is actually. 1/(5+3n). Now the integral test requires that the function be continuous, positive, and decreasing for x greater than or equal to 1. For x greater than or equal to 1, we can see that it is continuous and positive, now we need to just check the derivative to see if it is increasing or decreasing. So when I take the derivative via quotient or product rule, either one you prefer, I end up with -3/(5+3n)^2. So in order for the function to be decreasing, f'(x) has to be less than 0 for all x on the interval, which it will be. So given all that, now we just need to check the integral and see what happens. With me so far?

Answer 3

Thanks Psymon, I knew all that, its just trying to figure out the 1/(5+3n) is what i was having trouble with. Now that I know that I can proceed with the problem :)

Answer 4

Alright, np then, Ill leave the rest to you :3

Answer 5

how did you figure that out btw?

Answer 6

do you just play around with it or is there some kind of system?

Answer 7

Well, I did have to play with it a little to be honest.

Answer 8

Not always much of a system really, its kinda practice. Either way, I know I have to have a 5 somewhere in there, since it is my starting point in the denominator. From there, it's just finding or do I come up with the part that involves adding 3 to it each time. So I kinda played around with powers of 3, multiples of 3, and finally came to that.

Answer 9

oh ok, well that makes some sense now. thanks! I guess it just take much practice in order to be able to look at a problem and start seeing a pattern..

Answer 10

Unfortunately, yes. But when you do it long enough, you kinda have a better idea where to start so youre not wondering for as long. Good luck!

Answer 11

thanks!