find the sum of the series of (-2)^n/3^n+1.
This is an alternating series...
what if we rewrite it as
an= (1/3)* (-2/3)^(n)
Divide any nth term by the n-1 term and see if you get a ratio term r. This might be a geometric series.
To determine if the series is a geometric series, we need to find the common ratio, r.
We can do this by dividing any nth term by the (n-1)th term:
r = (-2/3)^(n) / (-2/3)^(n-1)
When we divide two powers with the same base, we subtract their exponents:
r = (-2/3)^(n) * (-3/2)^(n-1)
r = [(-2/3) * (-3/2)]^(n-1)
r = (1)^(n-1)
Since (1)^n = 1 for any value of n, the ratio, r, is equal to 1.
Now that we have determined the ratio, we can proceed to calculate the sum of the series.
The formula for the sum of an infinite geometric series is given by the formula:
S = a / (1 - r)
where a is the first term of the series and r is the common ratio.
In our case, the first term a is (1/3) * (-2/3)^(0) = 1/3.
Substituting the values into the formula, we have:
S = (1/3) / (1 - 1)
Since the denominator for the formula is (1 - r) and r is equal to 1, the denominator becomes 0. And dividing by 0 is undefined.
Therefore, the sum of the series is undefined.