Convergent Series
Find the sums of the series in Exercis...

Question

Convergent Series

Find the sums of the series in Exercises 19–24.

∑ (from n = 2 to ∞) -2/[n(n+1)]

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the sums of the series, the sum evaluated from n equals 3 to infinity of. 3 divided by n multiplied by the parentheses of n minus 1. OK, so it appears for this particular problem we're asked to ultimately solve for or to find the sums of the series. So what does s equal, and that is our final answer that we're ultimately trying to solve for. So we're trying to figure out what the sum of this specific series is. So with that in mind, based on our prior knowledge, we need to recall that in order to solve this problem, we will need to use the method of partial fraction decomposition in order to rewrite the general term as a telescoping sum. And by expressing the series as a sequence of partial sums, we will then be able to cancel out the middle terms, and that will help us find the limit of the remaining terms as n approaches infinity. So with that in mind, that's like a little blueprint for what we're going to do to solve this problem. Our first step that we need to take is we need to identify the general term of the series, and we need to note that the series is given to us as the sum evaluated from n equals 3 to infinity of 3 divided by n multiplied by parentheses of n minus 1. And we need to note that the general term is a subscript N. I'll call AN for short. AN is equal to 3 divided by N multiplied by parentheses of N minus 1. And once again, this is for our general term. So, moving right along to our second step, we now need to recall and use partial fraction decomposition. We need to note that we want to decompose the fraction into the following form, which states that 3 divided by n multiplied by parentheses of n minus 1 will be equal to drum roll. A divided by N plus B divided by N minus 1. So now, at this stage, we need to multiply both sides by N multiplied by parentheses of N minus 1. So, let's make a note of this. So, we're multiplying both sides, so I'll put a multiplication sign. We're going to multiply both sides by N multiplied by parentheses of N minus 1, and when we do that, we will get. 3, which will be equal to, so 3 is equal to a multiplied by parentheses of n minus 1 plus B multiplied by n. Now we need to solve for B, and in order to do that, we need to let n equals 1. So we plug in a value of 1 in for n, and we will get 3 is equal to b multiplied by 1, which will mean that b is simply equal to 3. Fantastic. And in order to find a, and let's make a note off to the side that how we solve for B is when we set n is equal to 1. And in order to solve for A, we need to let N equals 0. So let's plug in a value of 0 in for N. So 3 will be equal to a multiplied by 0 minus 1, which will mean when we rearrange this expression to isolate and solve for A, we will find that A is simply equal to -3. So with that in mind, the term will become 3 divided by N minus 1 minus 3 divided by N. Fantastic. Moving on to our third step, we now need to write out our partial sum S subscript K, which I'll call SK for short. And as we should recall, the partial sum SK will be the sum of the terms from, and let's make a note, from N equals 3 to N equals K. So once again, the partial sum K will be the sum of the terms from N equals 3 to. N equals K, meaning, as we should recall, we could write that SK will be equal to the sum evaluate from N equals 3 to k of parentheses of 3 divided by n minus 1 minus 3 divided by n. We now at this point, need to list the first few terms and the last term. So with that in mind, we can go ahead and write that win, so let's make a note that when N is equal to 3, that will mean that 3 divided by 2 minus 3 divided by 3. And when n is equal to 4, that will mean that 3 divided by 3 minus 3 divided by 4. And when n is equal to 5, that will mean that 3 is divided by 4 minus 3 divided by 5. And when N is equal to K, that will mean that 3 divided by K minus 1 minus 3 divided by K. Fantastic. So what do we do now? Well, for our 4th step is we now need to simplify our telescoping sum, and based on our prior knowledge, we need to observe that the negative part of each term will cancel out with the positive part of the next term. So we can go ahead and write that sk will be equal to 3 divided by 2. -3 divided by 3 + 3 divided by 3 minus 3 divided by 4 plus 3 divided by 4 minus continuing on. For a er like a bunch of different variables, -3 divided by K minus 1, plus 3 divided by K minus 1. -3 divided by k. Fantastic. And we need to note that after all the middle terms cancel, only the first and last parts will remain, meaning we could write that sk will be simply equal to 3 divided by 2 minus 3 divided by k. Fantastic. So our 5th and final step is we need to calculate the sum S by taking the limit. And we need to note, based on our prior knowledge, the sum of the infinite series will be the limit of the partial sum as K approaches infinity. So we could write that S is equal to the limit as K approaches infinity of parentheses of 3 divided by 2 minus 3 divided by K. And we need to note that as K becomes infinitely large, we need to note that 3 divided by K will approach 0, and 3 divided by K will approach 0 once again as K becomes infinitely large. So now we can evaluate and solve for our final answer. So you'll note that the sums is equal to 3 divided by 2 minus 0, which will be simply equal to 3 divided by 2. And that's our final answer. Hooray, we did it. So going back to look at the prom itself, we can safely say that our final answer once again without a doubt for the sum of this specific series will be that S is equal to 3 divided by 2, and that's it. That's our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Convergent SeriesFind the sums of the series in Exercises 19–24.∑ (from n = 2 to ∞) -2/[n(n+1)]

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Convergent Series
Find the sums of the series in Exercises 19–24.

∑ (from n = 2 to ∞) -2/[n(n+1)]