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.

63. ∑ (k = 1 to ∞) 1 / ((k + p)(k + p + 1)), where p is a positive integer

Accepted Answer

Welcome back, everyone. Consider the telescoping series sigma from K equals 1 up to infinity of 1 divided by K + R +1, multiplied by K + R + 2 where R is a positive integer. Evaluate the series by finding the formula for the nth term of the sequence of partial sums SM. And by evaluating the limits and approaches infinity of sm. So, first of all, let's analyze the N term formula for our series. We can write 1 divided by K + R + 1. Which is multiplied by K + R + 2. What we're going to do is simply use partial fraction decomposition. To rewrite it as a divided by K + R + 1 + B divided by our second linear factor which is K + R + 2. Now if we use the common denominator K + R + 1. Multiplied by K + R + 2, we can rewrite the numerator as a multiplied by K + R + 2. Plus B multiplied by K + R + 1. And now we expanding the numerator, we get a multiplied by K plus a multiplied by R. Plus 2 A. Plus B multiplied by K plus B multiplied by R + B. And now if we factor out K we get. A plus B multiplied by K. so we can essentially label those terms that contain K. Those are AK and BK. And for the remaining terms, we can just rewrite them as AR. + 2 A + B + B, right? And now we can introduce a set of equations. We can show that A plus B is going to be zero since we don't have any K times in the numerator. And then we can also show that AR + 2 A plus BR + B is equal to 1 because those terms correspond to the constant. Now, let's go ahead and show that B is equal to negative A from the first equation. And we can substitute this expression into the 2nd 1. So we get AR + 2A. Plus BR, which is just negative AR, plus B, which is negative A, and this is equal to 1. So now we can simplify and show that. A minus AR, those terms cancel each other out, right? And we get A equals 1. Meaning B, which is negative A is equal to -1. So essentially we can rewrite our. Keith term as one divided by. K + R + 1. Plus B divided by, since B is -1, we can just say minus 1 divided by K + R + 2. So, this is the case term formula, and we're going to use it to write several terms in a sum SM. Let's go ahead and do that. We can write S and equals. We know that the initial index in this problem begins with K equals 1. So when K is equal to 1, we get 1 divided by. 1 + R + 1, we get R + 2 minus 1 divided by 1 + R + 2, so that's R + 3. Then we add the next terms when the index is equal to 2. So we get 1 divided by 2 + R + 1, that's R + 3 minus 1 divided by K + R + 2. So 2 + R + 2 gives us R + 4. Less 1 K is equal to 3, we get 1 divided by R + 4 minus 1 divided by R + 5, and so on. Until we reach the nth term. For the nth term we would have an expression of 1 divided by n plus R + 1 minus. 1 divided by N plus R + 2. Now when we consider the behavior of. This expression, we can notice that we can cancel out 1 divided by R + 3 and 1 divided by R + 3, followed by -1 divided by R + 4 and. 1 divided by R + 4 and so on. And we will always end up with our first term and our last term remaining in the sum. So we're gonna cancel out everything up until the final term, right? And we're going to leave the first term. So now that we have the expression for us and we can. Find the limit as n approaches infinity of sn which has an expression of 1 divided by r plus 2. Minus 1 divided by n plus R + 2. Let's notice that the 2nd term in this limit is going to approach 0 because we're dividing 1 by an infinitely large number and approaches infinity, so the denominator. Approaches infinity as well. And then we end up with 1 divided by R + 2, where R is a constant, so our final answer would be 1 divided by R + 2. Thank you for watching.

Key Concepts

Telescoping Series

Partial Sums and Their Limits

Partial Fraction Decomposition

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