Sum of infinite series

[jayz]

12) Find the sum of the infinite series: \(\displaystyle \displaystyle{ \sum_{n\, =\, 1}^{\infty}\, \left(\frac{5}{2^n}\, -\, \frac{1}{3^n}\right) }\)

Why is it for this question the solution is (5/2)/(1-1/2) - (1/3)/(1-1/3) which is 9/2.

Isn't this a geometric series with a=5 r=1/2 for first part and a=1 and r=1/3 for the second part?

 

[Ishuda]

12) Find the sum of the infinite series: \(\displaystyle \displaystyle{ \sum_{n\, =\, 1}^{\infty}\, \left(\frac{5}{2^n}\, -\, \frac{1}{3^n}\right) }\)

Why is it for this question the solution is (5/2)/(1-1/2) - (1/3)/(1-1/3) which is 9/2.

Isn't this a geometric series with a=5 r=1/2 for first part and a=1 and r=1/3 for the second part?

For the first series:

\(\displaystyle \displaystyle{ \sum_{n\, =\, 1}^{\infty}\, \frac{5}{2^n}\, =\, 5\, \times\, \sum_{n\,=\,1}^{\infty}\, \frac{1}{2^n}\, =\, (5) \left(\frac{1}{2}\right)\left( \sum_{n\, =\, 1}^{\infty}\,\frac{1}{2^{n-1}}\right)\, =\, \left(\frac{5}{2}\right) \left(\sum_{n\, =\, 0}^{\infty}\,\frac{1}{2^n}\right)\, =\, \left(\frac{5}{2}\right) \left(\frac{1}{1\, -\, \frac{1}{2}}\right) }\)

and similarly for the other series.

 

[jayz]

_{OK thank you so much for your help these past few days! I finally understand this now}