Find the sum of the series the series from n equals 1 to infinity of the quotient of negative 1 raised to the n minus 1 power and n squared correct to two decimal places.
Type your answer in the space below and give 2 decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.48). (15 points) Please explain. I need help!
I thought we already did this. Look at the sequence:
1/1^2 - 1/2^2 + 1/3^2 - 1/4^2 + ...
Look familiar?
To find the sum of the series, we need to evaluate each term and add them up. Let's break down the given series:
Series: ∑(-1)^(n-1) / n^2, from n = 1 to infinity
The first step is to consider the individual terms of the series. We observe that the numerator alternates between positive and negative 1, depending on the value of n.
First term: n = 1
(-1)^(1-1) / 1^2 = 1/1 = 1
Second term: n = 2
(-1)^(2-1) / 2^2 = -1/4 = -0.25
Third term: n = 3
(-1)^(3-1) / 3^2 = 1/9 ≈ 0.11
We can notice that the sign of the term alternates with each term, while the denominator increases as n^2.
Now, to find the sum, let's add up the terms. However, summing an infinite number of terms can be challenging, so we'll need to find a way to simplify the series.
We can use the concept of the "alternating series," which allows us to determine if a series converges or diverges by examining the absolute values of the terms. In this case, we have an alternating series because the numerator alternates between positive and negative values.
To determine if the series converges, we need to check if the absolute values of the terms approach zero as n approaches infinity. Let's find the absolute values of the terms.
First term: |1| = 1
Second term: |-0.25| = 0.25
Third term: |0.11| = 0.11
The absolute values of the terms are decreasing, and it seems like they will approach zero as n increases. Therefore, we can conclude that the series converges.
Now, we can use a formula to find the sum of the convergent alternating series, which is given by:
Sum = (-1)^(n-1) / n^2
To find a more compact expression for the series, let's group the terms together:
Sum = (1/1) - (1/4) + (1/9) - (1/16) + (1/25) - ...
We notice a pattern in the denominators: they are perfect squares in increasing order.
Sum = 1 - 1/4 + 1/9 - 1/16 + 1/25 - ...
We can rewrite it slightly differently by factoring out 1/n^2:
Sum = 1(1 - 1/4 + 1/9 - 1/16 + 1/25 - ...)
Now, this resembles a known series called the "alternating series of reciprocals of perfect squares." The formula for this series is:
Sum = π^2 / 12
Using this formula, we can find the sum of our series:
Sum = π^2 / 12 ≈ 0.822
Therefore, the sum of the given series correct to two decimal places is approximately 0.82.