Consider the infinite series of the form:

(+/-)3(+/-)1(+/-)(1/3)(+/-)(1/9)(+/-)(1/27)(+/-)...(+/-)(1/3^n)(+/-)...

(A) Find x and y from: x(</=)(+/-)3(+/-)1(+/-)(1/3)(+/-)...(</=)y.
(B) Can you choose the signs to make the series diverge?
(C) Can you choose the signs to make the series sum to 3.5?
(D)Can you choose the signs to make the series sum to 2.25?
(E) Can you choose the signs to make the series sum to (159/36)?

For (C), can you make 3 and 1 both positive and then the rest of the series negative, which would be a negative geometric series? A little shaky on that.

And for (D), is that an alternating series?

(A) To find x and y, we need to determine the range of possible values for the series terms.

The series can be rewritten as:
\[S = 3 - 1 + \dfrac{1}{3} - \dfrac{1}{9} + \dfrac{1}{27} - \dots + (-1)^n \dfrac{1}{3^n} + \dots \]

The terms alternate between positive and negative, with each term becoming smaller due to the geometric factor of \(\dfrac{1}{3}\).

First, let's look at the partial sums when the minus signs are chosen:

\[S_1 = 3 - 1 = 2 \]
\[S_2 = 3 - 1 + \dfrac{1}{3} = \dfrac{8}{3} \]
\[S_3 = 3 - 1 + \dfrac{1}{3} - \dfrac{1}{9} = \dfrac{26}{9} \]
\[S_4 = 3 - 1 + \dfrac{1}{3} - \dfrac{1}{9} + \dfrac{1}{27} = \dfrac{80}{27} \]
\[ \vdots \]

We notice that the partial sums seem to approach a certain value. Let's call this value S. The partial sums converge to S if the absolute value of the common ratio (in this case, \(\dfrac{1}{3}\)) is less than 1. In this case, \(\dfrac{1}{3} < 1\), so the series converges.

Now, let's look at the partial sums when the plus signs are chosen:

\[S'_1 = 3 + 1 = 4 \]
\[S'_2 = 3 + 1 + \dfrac{1}{3} = \dfrac{13}{3} \]
\[S'_3 = 3 + 1 + \dfrac{1}{3} + \dfrac{1}{9} = \dfrac{46}{9} \]
\[S'_4 = 3 + 1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} = \dfrac{121}{27} \]
\[ \vdots \]

Again, we notice that the partial sums seem to approach a certain value, which we'll call \(S'\). The partial sums converge to \(S'\) if the absolute value of the common ratio (in this case, \(\dfrac{1}{3}\)) is less than 1. Since \(\dfrac{1}{3} < 1\), this series also converges.

Therefore, the value of \(x\) can be considered as the sums of the series with alternating signs (the minus signs):
\[x = \dfrac{80}{27}\]

And the value of \(y\) can be considered as the sums of the series with all positive signs:
\[y = \dfrac{121}{27}\]

(B) To make the series diverge, we need to ensure that the partial sums become unbounded with each subsequent term. In other words, we would need to choose signs in a way that the series terms grow and do not diminish in magnitude. However, as we can see from the analysis above, both choices of signs lead to convergent series. Thus, it is not possible to make the series diverge.

(C) To make the series sum to 3.5, we need to choose the signs in a way that the partial sums eventually reach this value. Considering that we have an alternating series, we can manipulate the signs to achieve this.

One possible approach is to make the positive terms equal to 3 and 1 in the series:
\[S = 3 + 1 - \dfrac{1}{3} + \dfrac{1}{9} - \dfrac{1}{27} + \dots \]

Then, in order to make the series sum to 3.5, the negative terms should sum up to -0.5:
\[-\left(\dfrac{1}{3} - \dfrac{1}{9} + \dfrac{1}{27} - \dots\right) = -0.5\]

We can achieve this by choosing the negative terms to be the negative of the positive terms, i.e., \(-3\) and \(-1\):
\[S = 3 + 1 - \dfrac{1}{3} + \dfrac{1}{9} - \dfrac{1}{27} + \dots -3 + (-1) + \dfrac{1}{3} - \dfrac{1}{9} + \dfrac{1}{27} - \dots\]

In this case, the series sum would be 3.5.

(D) Similarly, to make the series sum to 2.25, we can choose the positive terms to be 2 and 1:
\[S = 2 + 1 - \dfrac{1}{3} + \dfrac{1}{9} - \dfrac{1}{27} + \dots \]

Then, the negative terms should sum up to -0.75:
\[-\left(\dfrac{1}{3} - \dfrac{1}{9} + \dfrac{1}{27} - \dots\right) = -0.75\]

This can be achieved by choosing the negative terms to be the negative of the positive terms, i.e., \(-2\) and \(-1\):
\[S = 2 + 1 - \dfrac{1}{3} + \dfrac{1}{9} - \dfrac{1}{27} + \dots -2 + (-1) + \dfrac{1}{3} - \dfrac{1}{9} + \dfrac{1}{27} - \dots\]

In this case, the series sum would be 2.25.

(E) To make the series sum to \( \dfrac{159}{36} \), we need to analyze the partial sums and choose the signs appropriately.

(A) To find x and y from the expression x(≤)±3±1(±)(1/3)(±)...(≤)y, we need to understand the pattern of the infinite series.

If we look at the terms in the series, we notice that each term is a power of 1/3, starting with 1 (which is 3^0) and decreasing with each subsequent term. The signs in front of each term alternate between positive and negative. This is what makes it an alternating series.

Let's focus on the terms: 3, 1, 1/3, 1/9, 1/27, ...

We can see that the series is a combination of a geometric series and an alternating series.

(B) Can the signs be chosen to make the series diverge?

No, the signs cannot be chosen to make the series diverge. By the Alternating Series Test, if the terms of an alternating series decrease in absolute value and approach zero, then the series converges. In this case, as the terms of the series approach zero as the powers of 1/3 increase, the series converges.

(C) Can the signs be chosen to make the series sum to 3.5?

Yes, the series can be made to sum to 3.5. Since the series is an alternating geometric series, the sum is given by the formula:

Sum = a / (1 - r)

where a is the first term and r is the common ratio.

In this case, the first term a is 3, and the common ratio r is -1/3. Plugging these values into the formula:

Sum = 3 / (1 - (-1/3)) = 3 / (4/3) = 9/4 = 2.25

Therefore, choosing the signs accordingly, the series can be made to sum to 2.25.

(D) Can the signs be chosen to make the series sum to 2.25?

Yes, the signs can be chosen to make the series sum to 2.25. As explained in the previous answer, the series can be made to sum to 2.25 by choosing the appropriate signs.

(E) Can the signs be chosen to make the series sum to 159/36?

No, the signs cannot be chosen to make the series sum to 159/36. Since the series is an alternating geometric series, the sum can only be a rational number if the common ratio is between -1 and 1. In this case, the common ratio is 1/3, which is less than 1 but not between -1 and 1. Therefore, the series cannot sum to 159/36.