Are the first four terms of, (∞ Σn=2 (-2)^n-1) 1,-2,-4,-6? Does it Diverge? And is the sum 1/3?
This is just checking my answers!
The infinity symbol is on top of the summation notation and the n=2 is on the bottom of it. then the (-2)^(n-1) is right next to it. Please help!
for a series to converge,
|a_(n+1)/a_n| must be less than 1
clearly this series diverges
Okay, thanks! Would the sum of it be 1/3? I’m not really for sure on if it can even have a sum of 1/3 since the first four terms i got were 1,-2,-4,-6
since it diverges, it cannot have a sum!
You are a blessing! I was iffy about the whole thing. Thank you:)
To find the first four terms of the series (∞ Σn=2 (-2)^n-1), we can substitute the values of n from 2 to 5 into the series.
The given series is (∞ Σn=2 (-2)^n-1).
When n = 2:
(-2)^(2-1) = (-2)^1 = -2
When n = 3:
(-2)^(3-1) = (-2)^2 = 4
When n = 4:
(-2)^(4-1) = (-2)^3 = -8
When n = 5:
(-2)^(5-1) = (-2)^4 = 16
Therefore, the first four terms of the series are: -2, 4, -8, 16.
Now, let's determine if the series diverges. We can do this by observing the behavior of the terms of the series. In this case, the terms alternate between negative and positive values, becoming larger in absolute value as n increases.
Since the terms do not approach a specific value as n increases, and instead keep getting larger in magnitude, the series diverges.
Finally, let's find the sum of the series. In this case, the sum of the series can be calculated using the formula for the sum of an infinite geometric series.
The formula is:
Sum = a / (1 - r)
Where:
a represents the first term of the series
r represents the common ratio
In our case, a = -2 and r = -2.
Plugging in these values into the formula, we have:
Sum = -2 / (1 - (-2))
= -2 / (1 + 2)
= -2 / 3
So, the sum of the series is -2/3, not 1/3 as mentioned in your question.
To summarize:
The first four terms of the given series (∞ Σn=2 (-2)^n-1) are -2, 4, -8, 16. The series diverges because the terms do not approach a specific value. The sum of the series is -2/3, not 1/3.