Which one of the following series is convergent?
A. the summation from n equals 1 to infinity of 1 over the quantity n squared
B. the summation from n equals 1 to infinity of 1 over n raised the the one fourth power
C. the summation from n equals 1 to infinity of 1 over n raised to the negative one half power
D. the summation from n equals 1 to infinity of n squared plus 1 over n
sum 1/n^k converges for k>1
so, A
D = sum(n + 1/n), so both terms diverge.
lets gooo
To determine whether a series is convergent or divergent, we can use several different tests. In this case, we will consider the given options and discuss the appropriate tests for each.
A. The series ∑(1/n^2) is a well-known series called the "p-series." For the p-series, the general term of the series is in the form (∑(1/n^p)). The key to determining whether the series converges or diverges lies in the value of p.
For the given series, p = 2, which means it falls into the category of the p-series with p > 1. The p-series with p > 1 converges. Therefore, the series A, i.e., ∑(1/n^2), is convergent.
B. The series ∑(1/n^(1/4)) has the form of a power series, specifically with a fractional exponent. To determine whether this series converges or diverges, we can use the integral test. The integral test states that if a series, ∑(f(n)), is convergent if and only if the corresponding integral, ∫(f(x))dx, is convergent.
In this case, we have ∑(1/n^(1/4)) and its corresponding integral is ∫(1/x^(1/4))dx. Evaluating this integral, we get 4x^(3/4)/3 + C.
Now, let's look at the limits as x approaches infinity and as x approaches 1. As x approaches infinity, the integral becomes (∞)^(3/4) - (1)^(3/4) which equals ∞. Since the integral diverges, the series B, i.e., ∑(1/n^(1/4)), diverges as well.
C. The series ∑(1/n^(-1/2)) has a negative exponent. For this type of series, we can use the p-series test as explained in option A.
The value of p here is -1/2, which is less than 1. Therefore, the series C, i.e., ∑(1/n^(-1/2)), diverges.
D. The series ∑((n^2 + 1) / n) can be simplified as ∑(n + 1/n).
Since this series combines two terms, n and 1/n, we need to consider their convergence separately. The series ∑n diverges because it is a series of positive integers. However, the series ∑(1/n) is a harmonic series, which is known to diverge.
Since the series D, i.e., ∑((n^2 + 1) / n), contains both a divergent series (n) and a divergent series (1/n), it will also diverge.
In summary, among the given options, the only convergent series is A, i.e., ∑(1/n^2).