30. b. By differentiating the series in part (a) term by term,...

Question

30. b. By differentiating the series in part (a) term by term, show that

Σ(from n=1 to ∞) n / (n + 1)! = 1.

Accepted Answer

Consider the function F X equals 1 minus cosine x divided by x2. By finding the power series representation of FX centered at x equals 0 and differentiating it term by term, determine the value of the infinite series from 1 to infinity of -1 to the n + 1 multiplied by 2 n divided by 2 n + 2 factorial. Now, this first says define the power series representation, so we can do this by first looking at cosine x. Cosine x is one of our known power series, which is given by the sum. From 0 to infinity. Of -1 to the K. X 2K divided by 2K factorial. Which expands to 1 minus X2 divided by 2 factorial. Plus x 4 divided by 4 factorial. Minus x 6 divided by 6 factorial and so on. Now, we have 1 minus cosine x in our numerator, so you will subtract 1. We'll actually put it in the front and subtract cosine x from one. To give us our new series. Of X2 divided by 2 factor minus X4 divided by 4 factorial. Plus x 6 divided by 6 factorial and so on. The one in the previous series went away, and we'll just flip the signs for the rest of our terms. This actually simplifies to a different series. From 0 to infinity. Of 1 to the end. 2 to the end. Plus 2 Divided by 2n + 2 factorial. And we actually have an X there, instead of 2 to the end + 2. X to the 2 N + 2 is how we should write that. Now we divide all of our terms by X2 to find our correct power series for FX. So every term divided by x2 gives us 1 divided by 2 factorial. Minus X2 divided by 4 factorial. Plus x 4th divided by 6 factorial and so on. Which gives us our new series, being the sun. From 0 to infinity of -1 to the end. X 2 N divided by 2N plus 2. Factoria. Now, we're going to differentiate this term by term. And differentiate with respect to X. Now, our first term of this is actually a constant, because we noticed we have 1 divided by 2 factorial, that would go to 0. So then We're going to shift our index to start at 1, which means we'll take the derivative of this, F. Equals the sum from 1 to infinity. We'll put N equals in the front there. Now we just take the derivative in terms of X. This gives us -1 ton 2 N. Which comes in front Multiplied by x 2 n minus 1. Divided by 2n + 2 factorial. And now we'll evaluate this at x equals 1. We can then note that our main series we're looking for is S equals F. Of one This is because we have -1 to the end. In our target series. Is going to be -1 to the n + 1. This is based on what we're given in the very start of the problem. So we can write this ass equals F1 by removing that negative from our given series. So now we have S equals F1. So from here we can find the derivative of FFX. For this, we would use the quotient rule. We'll write DD X of. 1 minus cosine x divided by X2. Now this uses the quotient rule. Which will give us X 2 multiplied by sin x. 1 minus cosine x. Multiplied by 2 x. All divided by X22. This then simplifies. This gives us. X Sign X -2. Plus 2 cosine X. All divided by x to the 3rd. From here, we just substitute in our value X equals 1 to find F1. F1 then. Will be sign of one. -2 Plus 2 cosine of 1. And finally, since our approximation for S is negative F1. We'll just take the negative of this value. Which is 2 minus 2. Cosine of 1 minus sine of 1. This is a decimal number, so we can just leave it as this form. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

30. b. By differentiating the series in part (a) term by term, show thatΣ(from n=1 to ∞) n / (n + 1)! = 1.

Watch next

30. b. By differentiating the series in part (a) term by term, show that
Σ(from n=1 to ∞) n / (n + 1)! = 1.