Exercise 2.7.4. Give an example of each or explain why the request is impossible referencing the proper theorem(s).

(a) Two series \(\sum x_n\) and \(\sum y_n\) that both diverge but where \(\sum x_n y_n\) converges.

Let \(x_n = (0,1,0,1,...)\) and \(y_n = (1,0,1,0,....)\).

Then the sequence \(x_n * y_n = (0,0,0,0,...)\) and thus the sum of zeroes is zero.

(b) A convergent series \(\sum x_n\) and a bounded sequence \((y_n)\) such that \(\sum x_n y_n\) diverges.

Let \(x_n = (-1)^n (\frac{1}{n})\). From the alternating Series test we know that this is sequence will converge when we take the infinite sum.

Recall the Alternating Series Test states that if \((a_n)\) is a sequence satisfying,

1. \(a_1 \geq a_2 \geq a_3 \geq ... \geq a_n \geq ...\)

2. \((a_n) \rightarrow 0\)

Then, the alternating series \(\sum (-1)^{n+1}a_n\) converges.

Next, Let \(y_n = (-1)^n\). We know that y_n is going to be bounded by (\(|y_n| \leq 1\)).

So when we take the product of these two series we get \(\sum x_ny_n = \sum \frac{1}{n}\). Note that the right side is the harmonic series which we know diverges as a series.

(c) Two sequences \((x_n)\) and \((y_n)\) where \(\sum x_n\) and \(\sum(x_n + y_n)\) both converge but \(\sum y_n\) diverges.

That is, \(\sum x_n = X\) and

\(\sum(x_n+ y_n) = X+Y \leftarrow\) By the Algebraic Limit Theorem for Series.

This implies that \(\sum(x_n + y_n - x_n)=X + Y - X = Y = \sum y_n\). Thus by the algebraic limit Theorem for Series \(\sum y_n\) has to converge.

Algebraic Limit Theorem for Series: If \(\sum a_k = A\) and \(\sum b_k = B\), then

1. \(\sum(c a_k) = cA\) \(\forall c \epsilon \mathbb{R}\)

2. \(\sum(a_k +b_k) = A + B\)

(d) A sequence \((x_n)\) satisfying \(0 ≤ x_n ≤ \frac{1}{n}\) where \(\sum(−1)^n x_n\) diverges.

\(x_n = \left[ \begin{array}{lcr} \frac{1}{n} & if & n = 4m \mbox{ or } 4m+2 \\ \frac{1}{4m+2} & if & n=4m+1,i.e. 1,5,9,\mbox{...} \\ \frac{a}{2(4m+4)} & if & n=4m+3,i.e.3,7,11,\mbox{...}\end{array} \right]\).

If we write out the terms in the denominator we find that \[x_n^{-1} = (2,2,8,4,6,6,16,8,10,10,22,12,14,14,32,16,...)\] Then by taking \(\sum (-1)^nx_n\) we are get a sequence of the form \[(\frac{1}{4},\frac{1}{12},\frac{1}{20},...)= \frac{1}{4+8n}\]

Note that the Comparison Test says that if \(a_k\) and \(b_k\) are sequences satisfying \(0 \leq a_k \leq b_k\) for all \(k \epsilon \mathbb{N}\).

1. if \(\sum b_k\) converges, then \(\sum a_k\) converges.

2. if \(\sum a_k\) diverges, the \(\sum b_k\) diverges.

Note that a sequence less then the one we have above is \(a_n = \frac{1}{8n}\) for all \(n\geq1\). Also note that \(\sum a_n\) is a p series with p = 1. We know that this p series will then diverge. And according to the Comparison test So will our defined series \(x_n\).

Why not the Alternating Series Test?

Recall the Alternating Series Test that states if \((a_n)\) is a sequence satisfying,

1. \(a_1 \geq a_2 \geq a_3 \geq ... \geq a_n \geq ...\)

2. \((a_n) \rightarrow 0\)

Then, the alternating series \(\sum (-1)^{n+1}a_n\) converges.

Because in our question we were told to find a \(x_n\) satisfying \(0 ≤ x_n ≤ \frac{1}{n}\) which is not equivalent to \(x_1 \geq x_2 \geq x_3 \geq ... \geq x_n \geq ...\).