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.

57. ∑ (k = 1 to ∞) 1 / ((k + 6)(k + 7))

Accepted Answer

Welcome back, everyone. Consider the telescoping series sigma from K equals 1 up to infinity of 1 divided by K + 1 multiplied by K+2. Evaluate the series by finding the formula for the nth term of the sequence of partial sums SN and by evaluating the limit as N approaches infinity of SN. So, first of all, we're going to analyze the nth term of the sequence. Let's consider the original series in the form of 1 divided by K + 1 multiplied by K + 2. What we're going to do is simply use partial fraction decomposition and rewrite it in a form of A divided by K + 1 plus B divided by K + 2. If we rewrite this expression using the common denominator in terms of coefficients A and B. We're going to begin with the common denominator, and in the numerator, we're going to get A multiplied by K + 2 + B multiplied by K + 1. Expanding the numerator, we get a multiplied by K + 2 A. Plus B multiplied by K + B. And now factoring out, we got A plus B multiplied by K + 2 A plus B. Notice that the original numerator only has a constant, so A plus B is going to be equal to 0 since we don't have any key terms. And 2 A plus B is going to be equal to the constant, which is 1. We can now show that. B is equal to negative a from the first equation, and substituting into the second equation we get to a minus A equals 1. Meaning A is equal to 1 and B is negative A. Which is -1. So now Our terms. Correspond To the expression 1 divided by K + 1 multiplied by K + 2. Which is equal to a divided by K + 1, so that's 1 divided by K + 1 plus B divided by K + 2. So that's -1 divided by K + 2, because B is -1. And now if we consider our sum as N, it can be written as 1 divided by 1 + 1k begins with 1, so we got 11/2 minus 1/3 for K equals 1. Plus For k equals 2 we get 1/3 minus 1/4. Followed by for k equals 3 we get 1/4 minus 1/5. And so on until we reach our. Nth term, right? And the nth term is going to have 1 divided by N + 1 minus 1 divided by N + 2. So now let's not that we can cancel out 1/3 and 1/3, 1/4 and 1/4. Right, so we can cancel out all of those terms, and we always have our first term, 1/2, and the final term. So we can cancel out everything except from 1/2 and -1 divided by n + 2. So we now have our sum formula, and if we evaluate the limit as N approaches infinity of 1/2 minus 1 divided by N + 2. We get one half since 1 divided by an infinitely large number is going to approach 0, so the final answer for this problem is 1/2. 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.

57. ∑ (k = 1 to ∞) 1 / ((k + 6)(k + 7))

Key Concepts

Telescoping Series

Partial Fraction Decomposition

Limit of Partial Sums and Convergence

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