The ratio test to solve

The series in the problem is `sum_(n=1)^oo (2 +(-1)^n)/1.25^n`

To determine the convergence of the series let us use the comparison test. If there are two series `S_a` and `S_b` with terms `T_a` and `T_b` and `T_b >= T_a` , if the series `S_b` converges, `S_a` also converges.

Now, `(-1)^n` can be either positive or negative and this is dependent on whether n is even or odd.

Let `T_b = (2 + 1^n)/1.25^n` . If...

The series in the problem is `sum_(n=1)^oo (2 +(-1)^n)/1.25^n`

To determine the convergence of the series let us use the comparison test. If there are two series `S_a` and `S_b` with terms `T_a` and `T_b` and `T_b >= T_a` , if the series `S_b` converges, `S_a` also converges.

Now, `(-1)^n` can be either positive or negative and this is dependent on whether n is even or odd.

Let `T_b = (2 + 1^n)/1.25^n` . If `T_a = (2 +(-1)^n)/1.25^n` , `T_b >= T_a`

To determine the convergence of `S_b` we use the ratio test.

First we find the value of L = `lim_(n=oo) |T_(n+1)/T_n|`

`T_(n+1) = (2 +(1)^(n+1))/1.25^(n+1)`

= `(2 + 1^n)/(1.25*1.25^n)`

L = `lim_(n=oo) |T_(n+1)/T_n|`

= `lim_(n=oo)|((2 + 1^n)/(1.25*1.25^n))/((2 +1^n)/1.25^n)|`

= `1/1.25`

As 1.25 is greater than 1, `1/1.25` is less than 1.

By the ratio test, the series `S_b = sum(2 + 1^n)/1.25^n` converges.

As `S_b` converges, the series `S_a = sum (2 +(-1)^n)/1.25^n` also converges.

The given series `sum_(n=1)^oo (2 +(-1)^n)/1.25^n` converges.