Show that each series is a telescoping series. For each...

Key Concepts

Telescoping Series

A telescoping series is a series in which subsequent terms cancel out previous terms when the partial sums are computed. This cancellation causes most internal terms to vanish, leaving only a few terms from the beginning and the end, thus simplifying the evaluation of the series’ sum.

Partial Sums

Partial sums are the sums of the first n terms of a series and play a crucial role in the analysis of series convergence. In telescoping series, writing out the partial sums explicitly often makes evident the cancellation pattern that leads to a simple expression, facilitating the calculation of the series limit.

Convergence and Limits of Series

The convergence of a series is determined by taking the limit of the sequence of its partial sums as n approaches infinity. If the limit exists and is finite, the series is said to converge to that value. This principle is essential when evaluating series like telescoping series where the cancellation reduces the problem to evaluating the limit of a simplified expression.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to handle. This technique is useful in revealing the telescoping nature of a series by decomposing its general term into parts that cancel out in the sum.

Transcript

00:01 In this problem, we have given a telescoping series which is here sigma k is equals to 1 to infinity and this is 1 divided with k is multiplied with k plus 1 is multiplied with k plus 2.

00:16 And now we have to provide the general term s n and then we have to assign is which is the parcel sum of the series and then we have to determine the sum of the series if it converges.

00:30 So first step is to write this term as a sum of partial term so this can be written as say this would be 1 divided with this is k multiplied with k plus 1 multiplied with k plus 2 is equals to this is a divided with k plus b divided with k plus 1 plus c divided with k plus 2.

00:56 Now we would take the term so this would be here say 1 divided with k is multiplied with k plus 1 is multiplied with k plus 2 is equals to this is in the denominator this would be here a is multiplied with k plus 1 is multiplied with k plus 2 b is multiplied with k and this would be k and k plus 2 so this is k and k plus 2 and now c is multiplied with so k and k plus 2 so this is k and k plus 1 actually so k and k plus 1 is whole multiplied with say this is k multiplied with k plus 1 is multiplied with k plus 2 now this term this term and this term is cancelled out here and we left with this age 1 is equal to this is a multiplied with k plus 1 and k plus 2 is equal to k square plus 3k plus 2 and b is multiplied with k 2k plus c is multiplied with k squared plus 1.

02:19 Now segregate all the terms.

02:22 So this would be k square.

02:25 So k is square.

02:27 Say this would be a plus b plus c.

02:32 Now k so this would be k so this would be 3a so this is 3a and here this would be plus 2b so this is plus 2b and now the constant term so this is 2a so 2a plus c now see the coefficient so a plus b plus c is equal to 0 so we can say here a plus b plus c is equal to 0 equation number 1 second equation is 3a plus 2b is equals to again 0 and the third is 2a plus c is equals to we have 1.

03:14 Now we have 3 unknown variable and 3 equation so we have to find the value of a, b and c respectively.

03:23 So first we are finding the value of c.

03:24 C is equals to 1 minus 2a.

03:28 Now we have to put this value here so this would be a plus b.

03:33 Plus 1 minus 2a is equal to 0 now when we solve it so this would be b minus a so this is b minus a is equals to minus 1 and here we have say this is or we can say this is b is equals to this would be a minus 1 now we can put this value of b in this equation so this would be 3a plus 2 multiplied with a minus 1 is equal to 0.

04:07 So this would be 3a plus 2 a is equals to 5a...

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