true of false
if the sum of asubn from n=1 to infinity converges, and if a is not equal to 0, the the sum of 1/(a sub n) as n goes from 1 to infinity diverges.
True.
In order for the a subn series to converge, an infinite number of terms past some n value must be <0 , individually. That means there must be an infinite number of terms >0 in the 1/asubn series. Therefore it cannot converge.
The statement is true.
To understand why, let's consider the given situation.
Suppose we have a series, Σ(asubn) where n ranges from 1 to infinity, and it converges to some finite value. Also, assume that a is a non-zero constant.
Now, we want to determine whether the series Σ(1/(asubn)) diverges or not when n ranges from 1 to infinity.
To analyze this, we can make use of a concept called the convergence test. In particular, we will use the comparison test.
First, note that since the series Σ(asubn) converges, it implies that the terms asubn must tend to zero as n approaches infinity. (This is because if the terms didn't approach zero, the series would not converge.)
Now, using the comparison test, we can compare the series Σ(1/(asubn)) with Σ(1/n) where n ranges from 1 to infinity. Let's call this new series B.
The series B, Σ(1/n), is a well-known series named the harmonic series. It is well-known that the harmonic series diverges, meaning it does not converge to a finite value.
Now, if we compare the series Σ(1/(asubn)) with B, we can see that as long as asubn tends to zero, the series Σ(1/(asubn)) will behave like the harmonic series. Consequently, it will also diverge.
Therefore, we can conclude that if the series Σ(asubn) converges (with a ≠ 0), then the series Σ(1/(asubn)) diverges.
In summary:
- If Σ(asubn) converges (with a ≠ 0), then asubn tends to zero as n approaches infinity.
- Using the comparison test, we can compare Σ(1/(asubn)) with Σ(1/n).
- Since the series Σ(1/n) (harmonic series) diverges, and Σ(1/(asubn)) behaves similarly, it also diverges.