Finding sum of convergent series.

[peripatein]

Hi,

I have determined, correctly I believe, that the following series converges:
1/[(3n-2)(3n+1)]
Now I am asked to determine its sum. I have tried separating it into two subseries, but each time got a p-series with p=1, hence to no avail.
The answer should be 1/3, but how may it be arrived at?

 

[Mark44]

Hi,

I have determined, correctly I believe, that the following series converges:
1/[(3n-2)(3n+1)]
Now I am asked to determine its sum. I have tried separating it into two subseries, but each time got a p-series with p=1, hence to no avail.
The answer should be 1/3, but how may it be arrived at?

I would rewrite 1/[(3n-2)(3n+1)] as the sum of two fractions, using partial fraction decomposition. When you have done that, expand the new series. You'll probably find that the series is a telescoping one.

 

[Curious3141]

Hi,

I have determined, correctly I believe, that the following series converges:
1/[(3n-2)(3n+1)]
Now I am asked to determine its sum. I have tried separating it into two subseries, but each time got a p-series with p=1, hence to no avail.
The answer should be 1/3, but how may it be arrived at?

Partial fractions, following which you get a very simple telescoping series. Write out the first few terms of each.

 

[STEMucator]

EDIT : Whoops i didn't read the whole post, my bad.

 

[Mark44]

An easier way would be to use the comparison test. You should see the denominator is easily dominated by n²

The OP has already determined that the series converges. Now he/she needs to determine its sum.

 

[peripatein]

Thanks!
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibnitz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)

 

[SammyS]

Thanks!
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibnitz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)

I suggest starting a new thread for this.