# Series problem (again!)

• Jul 28th 2012, 01:10 PM
nisbahmumtaz
Series problem (again!)
Hi guys, so i was doing this one question and cant find a way through. It goes by like this :
What is the condition such that the series

1+[7/(3x+5)]+[7+(3x+5)^2].... is converge
hence show that the sum of the above series is 3x-5/3x-12
• Jul 28th 2012, 01:51 PM
emakarov
Re: Series problem (again!)
Quote:

Originally Posted by nisbahmumtaz
1+[7/(3x+5)]+[7+(3x+5)^2].... is converge
hence show that the sum of the above series is 3x-5/3x-12

Could you write the series and the answer correctly? I suspect the answer is not $3x-\frac{5}{3x}-12$.
• Jul 28th 2012, 01:57 PM
nisbahmumtaz
Re: Series problem (again!)
• Jul 28th 2012, 02:05 PM
Plato
Re: Series problem (again!)
Quote:

Originally Posted by nisbahmumtaz
Hi guys, so i was doing this one question and cant find a way through. It goes by like this :
What is the condition such that the series
1+[7/(3x+5)]+[7+(3x+5)^2].... is converge
hence show that the sum of the above series is 3x-5/3x-12

Is that + correct? Could it be $\frac{7}{(3x+5)}+\frac{7}{(3x+5)^2}+\frac{7}{(3x+5 )^3}\cdots~?$
If not then it diverges.
• Jul 28th 2012, 02:16 PM
emakarov
Re: Series problem (again!)
And $1+\sum_{i=1}^\infty\frac{7}{(3x+5)^i}=\frac{3x+11} {3x+4}$ (using geometric progression).

Edit: added ^i to the formula.
• Jul 28th 2012, 02:19 PM
nisbahmumtaz
Re: Series problem (again!)
Quote:

Originally Posted by Plato
Is that + correct? Could it be $\frac{7}{(3x+5)}+\frac{7}{(3x+5)^2}+\frac{7}{(3x+5 )^3}\cdots~?$
If not then it diverges.

Yikes. Darn it this iphone keyboard. Its supposed to be " / " not "+"
• Jul 28th 2012, 02:25 PM
Plato
Re: Series problem (again!)
Quote:

Originally Posted by nisbahmumtaz
Yikes. Darn it this iphone keyboard. Its supposed to be " / " not "+"