State whether this infinite series converges or diverges?
1+(t) + (t^3)+......
t=[(5x+6)/(3x-2)]
My thoughts on the question:
The sum of 'n' terms in a geometric progression is a[r^n - 1]/(r-1)--(let's calk this 1), where r>1 and when r<1, it is [a(1-(r^n)/(1-r) ---(let's call this 2)
Here a=(t^0) , r=t .But how do we decide which formula should be used from either 1 or 2,and apply the n-->(infinity) ,to find the above series converges or diverges?
We can see that t>1 for x>(-4) and t<1 for x<(-4)
If t = (5x+6)/(3x-2) then
|t| < 1
if -4 < x < -1/2
So we have to apply the two different formulae for the t>1 and t<1?
nope. It's just that the series converges iff |t| < 1
If you look at the formula, it's the same whether t<1 or t>1. The signs of the numerator and denominator both change, but the result is the same.
I missed that! Thank you!
To determine whether the infinite series converges or diverges, we can use the formula for the sum of an infinite geometric series. However, to apply this formula, we need to make sure that the common ratio (r) of the series lies between -1 and 1.
In this case, we have t = [(5x + 6)/(3x - 2)]. We need to substitute this value of t into the formula for the sum of an infinite geometric series and check the value of r.
The formula for the sum of an infinite geometric series is as follows:
Sum = a / (1 - r)
where 'a' is the first term and 'r' is the common ratio.
In this case, a = 1 and r = t = [(5x + 6)/(3x - 2)].
We substitute these values into the formula:
Sum = 1 / (1 - [(5x + 6)/(3x - 2)])
Now, we simplify this expression:
Sum = 1 / (3x - 2 - 5x - 6)
Sum = 1 / (-2x - 8)
Sum = -1 / (2x + 8)
To determine whether the series converges or diverges, we need to check the value of the common ratio (r) = [(5x + 6)/(3x - 2)].
Since the common ratio (-1/(2x + 8)) depends on 'x', we cannot definitively say whether the series converges or diverges without knowing the range of 'x' values.
To determine the convergence or divergence of the series, you need to evaluate the range of 'x' values for which the common ratio lies between -1 and 1. If the common ratio is within this range, the series converges; otherwise, it diverges.