71. Evaluating an infinite series two ways
Evaluate the ...

Question

71. Evaluating an infinite series two ways

Evaluate the series

∑ (k = 1 to ∞) (4 / 3ᵏ – 4 / 3ᵏ⁺¹) two ways.

a. Use a telescoping series argument.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says to find the sum of the infinite series, the sum of J equal to 1 to infinity of the quantity of 6 divided by 5 J. minus 6 divided by 5 raises to the quantity of J plus 1 in quantity, using a telescoping series argument. And we're given 4 possible choices as our answers. For choice A, we have 3 divided by 5. For choice B, we have 6 divided by 25. Choice C, we have 4 divided by 7, and for choice D, we have 6 divided by 5. The first thing we want to do is look at the 1st 3 values of J and our sequence. So, for J equal to 1, we will have the quantity of 6 divided by 5 minus 6 divided by 5 squared in quantity. You'll have plus for J equal to 2, we'll have the quantity of 6 divided by 5 squared. Minus 6 divided by 5 cubed in quantity, and then 4 J equal to 3 will have plus the quantity of 6 divided by 5 cubed, minus 6 divided by 54th power in quantity. Now, if you look at just these 1st 3 values of J, we see that the second part of J equal to 1 cancels out with the first part of J equal to 2. And the same is true if we look at the second part of J equal to 2, and the first part of J equal to 3. So that means that the partial sum here is just going to be equal to the first part of the first term, minus the last part of the last term. And so, if we look at the partial sum, which is going to be the sum, Of J equal to 1 ton. Of the quantity of 6 divided by 5 rates to the power J, minus 6 divided by 5 raises to the quantity of J plus 1 in quantity, that this partial sum is just going to be equal to 6 divided by 5 minus 6 divided by. 5 raise to the quantity of n plus 1 in quantity where all the other terms in the middle have canceled out. So this here is a telescoping series since all the terms in the middle cancel out and the actual sum is just based of one part of the first term and part of the last term. So, in order to find the sum of our infinite series, we just need to take the limit. As N approaches infinity of the quantity of 6 divided by 5 minus 6 divided by 5 rates to the quantity of N + 1 in quantity. Now as it approaches infinity. 6 divided by 5 rates of the quantity of N + 1 in quantity is going to approach 0. So applying the limits, this is going to be equal to 6 divided by 5 minus 0, which is just equal to 6 divided by 5. So that is the answer that we're looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice D. So, I hope that this explanation has been useful, and I'll see you in the next video.

71. Evaluating an infinite series two ways
Evaluate the series
∑ (k = 1 to ∞) (4 / 3ᵏ – 4 / 3ᵏ⁺¹) two ways.
a. Use a telescoping series argument.

Key Concepts

Telescoping Series

Infinite Geometric Series

Partial Sums and Convergence

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