54–69. Telescoping series
For the following telescoping ...

Question

54–69. Telescoping series

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.

61. ∑ (k = 1 to ∞) ln((k + 1) / k)

Accepted Answer

Welcome back, everyone. Consider the telescoping series sigma from K equals 1 up to infinity of LN of K divided by K + 1 evaluated the series by finding the formula for the nth term of the sequence of partial sums SN and by evaluating the limits and approaches infinity of SM. A 1/2 B 5 divided by 11. C 4 divided by 7, and D the series diverges. So for this problem, let's analyze the nth term of the series. In particular, we can write that a subscript K is equal to LN of K divided by K + 1, and using the properties of logarithms we can write it as LN of K. Minus LN of K + 1. So what we're going to do is simply write our sum formula for N terms. SN can be written as, starting with K equals 1, we get LN 1 minus. L and of 2 because K is equal to 1 we get these terms. Plus When K is equal to 2, we get ln of 2 minus LN of 3. Plus when K is equal to 3 we get ln of 3 minus LN of 4 and so on. And so we reach our final term which would be. And And that final term is going to have a form of LN of N minus LN of N + 1. Because for the final value of Nk becomes m. If we observe the behavior of this series, notice that we can cancel out ne Alan of 2 and Alan of 2, negative of 3 and Alan of 3, and over time we will only have the first term and the final term in our expression. So what we can do is simply cancel out all of the terms except from the first term and the last term. And now that we have the nth term of the sequence of partial sums, we can evaluate the sum. In the form of sigma. From K equals 1 up to infinity of LN of K divided by K + 1 using the limit. As an approaches infinity. Of 1 of 1, our first term minus Ln of N + 1. Now, LN of 1 is equal to 0, so we can cancel out that term. And now as an approaches infinity, we get negative LN of an infinitely large number, right, which essentially gives us a limit of infinity. So in this problem, we have that negative sign. We're going to add it and write negative infinity. So, since we don't get a finite value, we can simply conclude that the series diverges. So the correct answer to this problem corresponds to the answer choice D. Thank you for watching.

54–69. Telescoping series
For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.

61. ∑ (k = 1 to ∞) ln((k + 1) / k)

Key Concepts

Telescoping Series

Partial Sums and Their Limits

Properties of Logarithms

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