Which of the following series is NOT absolutely convergent?
A. the summation from n equals 1 to infinity of the quotient of negative 1 raised to the n plus 1 power and n
B. the summation from n equals 1 to infinity of the quotient of the quantity negative 1 and 2 raised to the nth power
C. the summation from n equals 1 to infinity of the quotient of negative 1 raised to the n plus 1 power and n raised to the 3 halves power
D. the summation from n equals 1 to infinity of the quotient of negative 1 raised to the n plus 1 power and the quantity n squared plus 1
|A| is just the Harmonic Series, so it diverges
|B| is just a GP with r = 1/2, so it converges
|C| sum 1/k^n converges for n>1
|D| a_n < 1/n^2 so it converges
To determine which of the given series is not absolutely convergent, we need to check the convergence of each series individually.
For series A,
the term of the series can be written as (-1)^(n+1)/n.
Since the numerator alternates between -1 and 1, and the denominator increases to infinity, the terms of the series oscillate between positive and negative values as n increases.
To determine if this series converges, we can check the convergence of the absolute value of the terms by removing the alternating signs.
The absolute value of the terms is 1/n, which is the harmonic series.
The harmonic series diverges since the sum of 1/n does not approach a finite value.
Therefore, series A is NOT absolutely convergent.
For series B,
the term of the series can be written as (-1/2)^n.
The absolute value of the terms is (1/2)^n, which is a geometric series with a common ratio less than 1.
The geometric series with a common ratio less than 1 converges.
Therefore, series B is absolutely convergent.
For series C,
the term of the series can be written as (-1)^(n+1)/(n^(3/2)).
Similar to series A, the numerator alternates between -1 and 1, and the denominator increases to infinity.
To determine if this series converges, we can check the convergence of the absolute value of the terms by removing the alternating signs.
The absolute value of the terms is 1/(n^(3/2)).
Since the exponent of n is greater than 1, the series is convergent.
Therefore, series C is absolutely convergent.
For series D,
the term of the series can be written as (-1)^(n+1)/(n^2 + 1).
Similar to series A and C, the numerator alternates between -1 and 1, and the denominator is a quadratic function of n.
Again, we can check the convergence of the absolute value of the terms by removing the alternating signs.
The absolute value of the terms is 1/(n^2 + 1).
Since the denominator is a quadratic function of n, the series is convergent.
Therefore, series D is absolutely convergent.
In conclusion, the series that is NOT absolutely convergent is: A. the summation from n equals 1 to infinity of the quotient of negative 1 raised to the n plus 1 power and n.
To determine which series is NOT absolutely convergent, we need to analyze the convergence of each series. Here's how you can determine the convergence for each option:
A. The series is the summation from n = 1 to infinity of (-1)^(n+1)/n.
To check for absolute convergence, we look at the series obtained by taking the absolute value of each term: 1/n.
This is a well-known p-series with p = 1, and it is known to be divergent. Therefore, the series in option A is not absolutely convergent.
B. The series is the summation from n = 1 to infinity of (-1/2)^n.
Again, we check for absolute convergence by considering the series obtained by taking the absolute value of each term: (1/2)^n.
This is a geometric series with a common ratio between 0 and 1, which makes it absolutely convergent. Therefore, the series in option B is absolutely convergent.
C. The series is the summation from n = 1 to infinity of (-1)^(n+1)/(n^(3/2)).
We can again check for absolute convergence by considering the series obtained by taking the absolute value of each term: 1/(n^(3/2)).
This is another well-known p-series, but this time with p = 3/2. It is also known to be convergent. Therefore, the series in option C is absolutely convergent.
D. The series is the summation from n = 1 to infinity of (-1)^(n+1)/(n^2 + 1).
Once more, we check for absolute convergence by taking the absolute value of each term: 1/(n^2 + 1).
This series is a variation of the well-known p-series with p = 2. It is convergent. Therefore, the series in option D is absolutely convergent.
After analyzing all the options, we can conclude that the series in option A is the one that is NOT absolutely convergent.