Question

How do you determine if `Sigma (7*5^n)/6^n` from `n=[0,oo)` converge and find the sums when they exist?

Answer 1

`sum_(n=0)^oo (7*5^n)/6^n = 42`

Explanation:

Write the series as:

`sum_(n=0)^oo (7*5^n)/6^n = 7 sum_(n=0)^oo (5/6)^n`

so this is a geometric series of ratio `r=5/6`.

As `abs(r) < 1` the series is convergent and its sum is:

`7 sum_(n=0)^oo (5/6)^n =7/(1-5/6) = 7/(1/6) = 42`