Determine whether the following diverges or converges:
$\displaystyle \sum_{k=1}^{\infty}\frac{k!}{k^{k}}$
The ratio test is inconclusive, and it seems like I'll need to use some form of the comparison test, but I'm stuck on where to begin.
Determine whether the following diverges or converges:
$\displaystyle \sum_{k=1}^{\infty}\frac{k!}{k^{k}}$
The ratio test is inconclusive, and it seems like I'll need to use some form of the comparison test, but I'm stuck on where to begin.
For the ratio test I got:
$\displaystyle \lim_{k\to \infty}\frac{(k+1)k!}{k\cdot k^{k}}\cdot \frac{k^{k}}{k!}=\lim_{k\to \infty} \frac{k+1}{k}=1$
And I understand that $\displaystyle k! \le k^{k}$, but I'm not sure how to apply that using a comparison; I always end up setting up a comparison where the larger series is divergent, which doesn't help.
Use NoncomAlg observation
$\displaystyle \lim_{k \to \infty} \bigg|\frac{(k+1)!}{(k+1)^{k+1}}\cdot \frac{k^k}{k!} \bigg|$
$\displaystyle \lim_{k \to \infty} \bigg|\frac{(k+1)!}{(k+1)^{k}(k+1)}\cdot \frac{k^k}{k!} \bigg|$
$\displaystyle \lim_{k \to \infty} \bigg| \frac{k^k}{(k+1)^k}\bigg|$
$\displaystyle \lim_{k \to \infty} \bigg| \left( \frac{k}{k+1}\right)^k\bigg|=\lim_{k \to \infty} \bigg| \left( \frac{1}{1+\frac{1}{k}}\right)^k\bigg|=\frac{1}{e}$
$\displaystyle \sum_{k=1}^{\infty}\frac{k^{k}}{k!}$
$\displaystyle \lim_{k\to \infty}\frac{a_{k+1}}{a_{k}}=\lim_{k\to \infty}\frac{(k+1)^{k+1}}{(k+1)k!}\cdot \frac{k!}{k^{k}}=\lim_{k\to \infty}(\frac{k+1}{k})^{k} = e $
And since $\displaystyle e > 1$, by the ratio test, the series diverges.
I guess you can see that it diverges that since $\displaystyle k^{k}\ge k!$, the sequence is increasing, which means $\displaystyle \lim_{k\to \infty} a_{k} \ne 0$, and the series must therefore diverge.
my question was to find $\displaystyle \lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}.$ the point here's that $\displaystyle \lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right|=\lim_{k\to\infty} \sqrt[k]{|a_k|},$ if one of the limits exists of course. that's why the ratio test is inconclusive if and only if the root test is
inconclusive. now let $\displaystyle a_k=\frac{k^k}{k!}.$ you showed that $\displaystyle \lim_{k\to\infty} \frac{a_{k+1}}{a_k}=e.$ thus we must also have $\displaystyle e=\lim_{k\to\infty} \sqrt[k]{a_k}=\lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}.$
My mistake for misreading your question. Also, for some reason our Calculus 3 class omits teaching the root test, so I'll have to read up on that on my own when time permits. Right now, I have to focus on the material that's gonna be on my upcoming exam.
And I'll try that one another time, my brain is dead for the day.
Well, I'm pretty sure it'll diverge because if I recall correctly from class, since you can express that series as:
$\displaystyle \sum_{k=1}^{\infty}2^{k} -\sum_{k=1}^{\infty}1$
It is pretty obvious $\displaystyle \sum_{k=1}^{\infty}1$ will diverge, so the whole series diverges. Correct me if I'm wrong though.