42–76. Convergence or divergence Use a convergence test of you...

Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)k! / (eᵏ kᵏ)

Accepted Answer

Welcome back everyone. In this problem, does the series M factorial divided by 3 to the M multiplied by M to M between M equals 1 and infinity converge or diverge? A says it converges and B says it diverges. Now notice that in this series we have factorials and powers. So the ratio test would be a good idea here to figure out if the series converges or diverges. Recall that by the ratio test if we find a limit L as M approaches infinity. OK, of AM + 1 divided by AM, that is for a series AM with AM greater than 0. OK. Then if L is less than 1, the series converges absolutely. OK, let me write that properly here. If L is greater than 1, the series diverges. And if L equals one, the series or the test, sorry, is inconclusive inconclusive. So if we can figure out our limit L, then we should be able to tell if the series will converge or diverge. Now in comparison. Then let's let A be equal to our term for a series. M factorial divided by 3 to the M multiplied by M to the M. No, by the ratio test, that means we'll need 8 and plus 1. And that means we'll just replace M with M+1 in our expression. So A M + 1 is going to be equal to M+1 factorial. Divided by 3 M + 1 multiplied by M + 1 to the M + 1. So now in this case. That means I limit L, OK. is going to be equal to the limit as M approaches infinity, OK. Of the absolute value of A M + 1. So that's M + 1 factorial divided by 3 to the M + 1 multiplied by M + 1 to the M + 1. OK. Divided by a.m. So we can, that it, that's the same as multiplying by its reciprocal. So that's gonna be multiplied by 3 to the M, multiplied by M to the M divided by M factorial. Now let's try to simplify our expression, OK? Now here there are a few things that we could do, right? Now if we combine both of these terms, OK. We combine both of these terms. Then we're going to have M + 1. OK, multiplied by M to the M. M + 1 multiplied by M sorry, yes, M + 1 multiplied by M to the M. And in our denominator, we'll have 3 multiplied by M + 1 to the M + 1, OK? Now we can simplify this further because we know that M+1 raise to the power of M+1, OK, is really. M + 1 multiplied by M + 1 to the M. I know that helps because that will help us to cancelload M+1 in our numerator, OK? And in that case, OK, that means we could rewrite. Our expression the limit as M approaches infinity of M to the M divided by 3 multiplied by M + 1 to the M because here as we say, it would have canceled this out in our denominator, OK. Now, notice that both of these terms are raised to the M, so we could rewrite this, OK. As M Sorry, 1/3. OK. Of M Divided by M + 1, all to the M. OK? So no, this is our ratio. 1/3 of M divided by M + 1, all to the M. But no, notice here if we take out our constant, this is 1/3 of our limit as M approaches infinity of M divided by M + 1 to the M. Which we could rewrite as a third of our limit as M approaches infinity of 1 divided by 1 plus 1 divided by M to the M. If we divided through our entire term by M. Mono recall by the standard limit that this expression, OK, all of this expression is the reciprocal of e. So really this is going to be 1/3 multiplied by the reciprocal of E or 1 divided by 3 E. Does L OK. Oh. Is equal to 1 divided by 3E. Now how does, how does that help? Well, 1 divided by 3E is less than 1, so that means the series converges by the ratio test. Therefore, A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)k! / (eᵏ kᵏ)

Key Concepts

Infinite Series and Convergence

Ratio Test

Factorials and Exponential Growth

Watch next