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Numerade Educator



Problem 44 Hard Difficulty

For which positive integers $ k $ is the following series convergent?

$ \displaystyle \sum_{n = 1}^{\infty} \frac {(n!)^2}{(kn)!} $


Series is convergent when $\mathrm{k} \geq 2$


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Lio I.

July 30, 2020

What if (n!)^6/(kn)!

Video Transcript

let's use the ratio test here to determine which values of K make the given Siri's conversion. So here this will be our A. So for the ratio test, we know to look at the limit and goes to infinity absolute value and plus one over. And so we also weaken Drop the absolute value here because an and care positive So and plus one i'LL do this red This is and plus one factorial It's square that for the denominator just replaced that end with an A plus one and then we'LL go ahead and divide that. And now using blue were just right r a n and factorial square kay and factorial And now, as usual, when you're dividing a fraction by another, just go ahead and flip that blue fraction upside down. But also here let me just rewrite I'LL use the fact that in plus one factorial by definition it is the product of the first and plus one positive imagers. But if I group the first end, I could just rename that within factorial times and plus one. So I have in factorial times and plus one and that's being squared and the denominator i'll just leave that for right now. And then I flip this fraction upside down and multiply that and you could kind of see why I The reason for doing this over here is that it allows us to cancel this in factorial square with this and factorial squared, leaving us with the limit and goes to infinity. And then we'LL have n plus one squared up top and now in the bottom, we're looking at the fraction Kay and factorial over K and plus one factorial. Now, how to write each of these out using the definition the top is just a product of from one to all the way to Cannes. The denominator Well, Cayenne is less than K times and plus one So eventually I'll come across these terms and then I'LL keep going until eventually we reach can plus one. And by writing it this way, you see, you could cancel the first can and so that just leaves us with this in the denominator. Now, when we go ahead and take this limit, so K's a positive image or so the smallest K to consider would be K equals one. However, of K equals one, then in that case. Excuse me. Hear this? So this is going up to K n. If I'm also applied that out, this is kay And plus que So if k equals one, the last term is just the first term. So that just means there's only one term. So Kay is one. The limit is the limit of n plus one squared over and plus one. But that goes to infinity, which is bigger than one and so a diver just in that case, So we do not want K equals one. Now if K equals two, then we would have and plus one square over. So now we would have to an plus one to an plus two. And one could use Lopez's house rule here to see that this living goes toe one over fourth which is less than one. So K equals two will give convergence and similarly, here's the key step here. If K is bigger than two, then the denominator has degree k, which is also larger than to because you'LL be multiplying K terms here. Just like when Kei was too will have to terms. And if Kay was three, you'd have three terms since the denominator has larger degree than the numerator. That will mean that the limit of an plus one over a and zero, which is less than one and so by ratio by racial test accuse me. That means convergence. It's okay bigger than two or equal to two gives convergence, so we will go ahead and leave. Is our final answer K larger than or equal to two.