Finding the limit of a sequence: a_k = ( k! )^( 1/k )

[ksdhart]

I'm having difficulty with a problem in my Calculus III course. The problem text is as follows:

For the sequences in Exercises 53 and 54, determine whether the sequence converges or diverges. If the sequence converges, give the limit.

53) \(\displaystyle a_k=\left(k!\right)^{\frac{1}{k}}\)

My initial thought was that the sequence must converge to 1, because as k gets infinitely large, 1/k will approach 0 and thus the whole thing will approach k^0 or 1. However, the answer key says that the sequence diverges. I guess that's because k factorial somehow grows so unbelievably fast as to overcome the exponent going to 0? In any case, I tried another line of reasoning to reason out why the sequence might diverge. I know that for any b > 1 and sufficiently large k, the following holds true:

\(\displaystyle b^k<k!\)

So, then I believe I'm justified in making this next step:

\(\displaystyle \left(b^k\right)^{\frac{1}{k}}<\left(k!\right)^{{1} \over {k}}\)

\(\displaystyle b<\left(k!\right)^{\frac{1}{k}}\)

But I don't know where to go from here. Is that the entirety of my proof? Because as k approaches infinity, it will still be bigger than any arbitrary constant b I could possibly pick? I'm quite confused. Any help would be appreciated.

 

[Ishuda]

I'm having difficulty with a problem in my Calculus III course. The problem text is as follows:



My initial thought was that the sequence must converge to 1, because as k gets infinitely large, 1/k will approach 0 and thus the whole thing will approach k^0 or 1. However, the answer key says that the sequence diverges. I guess that's because k factorial somehow grows so unbelievably fast as to overcome the exponent going to 0? In any case, I tried another line of reasoning to reason out why the sequence might diverge. I know that for any b > 1 and sufficiently large k, the following holds true:

\(\displaystyle b^k<k!\)

So, then I believe I'm justified in making this next step:

\(\displaystyle \left(b^k\right)^{\frac{1}{k}}<\left(k!\right)^{{1} \over {k}}\)

\(\displaystyle b<\left(k!\right)^{\frac{1}{k}}\)

But I don't know where to go from here. Is that the entirety of my proof? Because as k approaches infinity, it will still be bigger than any arbitrary constant b I could possibly pick? I'm quite confused. Any help would be appreciated.

You are almost finished. If there is some K₀ such that for any b>1
\(\displaystyle b<\left(K_0!\right)^{\frac{1}{K_0}}\)
then that is true for all k>K₀.

Suppose the sequence does tend to a limit L. What does that mean? [Hint: For all epsilon greater then zero, ...] Now choose b=L+10 and epsilon equal to 1.

 

[ksdhart]

Ah. Thank you. I understand. I want to set up a proof by contradiction. And just when I thought I'd never use a delta-epsilon proof ever again in my life...