Find a series ∑a_n for which ∑(a_n)^2 converges but ∑|a_n| diverges
Consider
1 - 1/2 + 1/3 - 1/4 + ... (alternating harmonic series)
1 + 1/4 + 1/9 + 1/16 + ... = π2/6
1 + 1/2 + 1/3 + 1/4 + ... diverges (harmonic series)
thanks!!
To find a series that satisfies this condition, we can make use of the concept of alternating series. An alternating series is one in which the terms alternate in sign (positive and negative).
Consider the series ∑((-1)^n)/√n. This series alternates between positive and negative terms. To test whether the series ∑((-1)^n)/√n converges, we will check two conditions:
1. The sequence of terms, √n, approaches zero as n approaches infinity.
2. The terms of the series are decreasing in absolute value.
First, note that as n approaches infinity, the denominator √n goes to infinity. Consequently, the sequence √n approaches zero, meeting the first condition.
Next, we observe that the terms of the series, ((-1)^n)/√n, are alternately positive and negative. Taking the absolute value of each term gives us the series ∑1/√n, which is a p-series with p = 1/2. The p-series ∑(1/n^p) converges if p > 1 and diverges if p ≤ 1.
In this case, p = 1/2, which is less than 1. Therefore, the series ∑1/√n (or equivalently, ∑((-1)^n)/√n) diverges.
Now, let's consider the series ∑((((-1)^(n+1))/√n)^2). This is the square of the terms in the original series.
Since the original series ∑((-1)^n)/√n is alternating, its terms are decreasing in absolute value. Thus, when we square the terms, the resulting series ∑(((−1)^(n+1))^2/√n^2) is composed of positive terms.
By the Alternating Series Test, if the terms of an alternating series are decreasing in absolute value, then the series of squared terms also converges.
Therefore, by taking the original series ∑((-1)^n)/√n and squaring each term, we obtain a series ∑((((-1)^(n+1))/√n)^2) for which ∑((((-1)^(n+1))/√n)^2) converges but ∑|((-1)^n)/√n| diverges.