Yet another series question

**Pinkk** · May 9th 2009, 07:07 PM

Determine whether the following diverges or converges:

$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$

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.

**TheEmptySet** · May 9th 2009, 07:09 PM

Originally Posted by Pinkk

Determine whether the following diverges or converges:

$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$

The ratio test is inconclusive, and it seems like I'll need to use some of the comparison test, but I'm stuck one where to begin.

Note that for $k > 2$ that

$k! < k^k$

check the case k =4

$1\cdot 2\cdot 3 \cdot 4 < 4\cdot 4 \cdot 4 \cdot 4$

**NonCommAlg** · May 9th 2009, 07:14 PM

Originally Posted by Pinkk

Determine whether the following diverges or converges:

$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$

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.

actually the ratio test works very nicely. you'll get $\lim_{k\to\infty} \frac{a_{k+1}}{a_k} = \frac{1}{e} < 1,$ and so the series is convergent.

**Pinkk** · May 9th 2009, 07:20 PM

For the ratio test I got:

$\lim_{k\to \infty}\frac{(k+1)k!}{k\cdot k^{k}}\cdot \frac{k^{k}}{k!}=\lim_{k\to \infty} \frac{k+1}{k}=1$

And I understand that $k! \le k^{k}$ , 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.

**TheEmptySet** · May 9th 2009, 07:27 PM

Originally Posted by Pinkk

Determine whether the following diverges or converges:

$\sum_{k=1}^{\infty}\frac{k!}{k^{k}}$

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.

Use NoncomAlg observation

$\lim_{k \to \infty} \bigg|\frac{(k+1)!}{(k+1)^{k+1}}\cdot \frac{k^k}{k!} \bigg|$

$\lim_{k \to \infty} \bigg|\frac{(k+1)!}{(k+1)^{k}(k+1)}\cdot \frac{k^k}{k!} \bigg|$

$\lim_{k \to \infty} \bigg| \frac{k^k}{(k+1)^k}\bigg|$

$\lim_{k \to \infty} \bigg| \left( \frac{k}{k+1}\right)^k\bigg|=\lim_{k \to \infty} \bigg| \left( \frac{1}{1+\frac{1}{k}}\right)^k\bigg|=\frac{1}{e}$

**Pinkk** · May 9th 2009, 07:28 PM

Bah, I'm an idiot; forgot to change the base of $k^{k}$ . Thanks.

**NonCommAlg** · May 9th 2009, 07:38 PM

here's a related question: what is $\lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}$ ?

**Pinkk** · May 9th 2009, 07:49 PM

$\sum_{k=1}^{\infty}\frac{k^{k}}{k!}$

$\lim_{k\to \infty}\frac{a_{k+1}}{a_{k}}=\lim_{k\to \infty}\frac{(k+1)^{k+1}}{(k+1)k!}\cdot \frac{k!}{k^{k}}=\lim_{k\to \infty}(\frac{k+1}{k})^{k} = e$

And since $e > 1$ , by the ratio test, the series diverges.

I guess you can see that it diverges that since $k^{k}\ge k!$ , the sequence is increasing, which means $\lim_{k\to \infty} a_{k} \ne 0$ , and the series must therefore diverge.

**NonCommAlg** · May 9th 2009, 08:10 PM

Originally Posted by Pinkk

$\sum_{k=1}^{\infty}\frac{k^{k}}{k!}$

$\lim_{k\to \infty}\frac{a_{k+1}}{a_{k}}=\lim_{k\to \infty}\frac{(k+1)^{k+1}}{(k+1)k!}\cdot \frac{k!}{k^{k}}=\lim_{k\to \infty}(\frac{k+1}{k})^{k} = e$

And since $e > 1$ , by the ratio test, the series diverges.

I guess you can see that it diverges that since $k^{k}\ge k!$ , the sequence is increasing, which means $\lim_{k\to \infty} a_{k} \ne 0$ , and the series must therefore diverge.

my question was to find $\lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}.$ the point here's that $\lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right|=\lim_{k\to\infty} \sqrt[k]{|a_k|},$ 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 $a_k=\frac{k^k}{k!}.$ you showed that $\lim_{k\to\infty} \frac{a_{k+1}}{a_k}=e.$ thus we must also have $e=\lim_{k\to\infty} \sqrt[k]{a_k}=\lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}.$

**Banned for attempted hacking** · May 9th 2009, 08:11 PM

You seem to like series. Try this one

$<br /> \sum_{k=1}^{\infty} 2^{n} -1<br />$

**Pinkk** · May 9th 2009, 08:16 PM

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.

**Banned for attempted hacking** · May 9th 2009, 08:19 PM

Give it a try sometime.

It's a sneaky series that find it's way into my first test on infinite series last semester. It took 15 minutes for me to figure it out.

**Pinkk** · May 9th 2009, 08:26 PM

Well, I'm pretty sure it'll diverge because if I recall correctly from class, since you can express that series as:

$\sum_{k=1}^{\infty}2^{k} -\sum_{k=1}^{\infty}1$

It is pretty obvious $\sum_{k=1}^{\infty}1$ will diverge, so the whole series diverges. Correct me if I'm wrong though.

**simplependulum** · May 9th 2009, 09:11 PM

It is necessary to use ratio test if the given sequence is a decreasing sequence so i d wish to know how to prove k!/k^k is decreasing everywhere ......

**Banned for attempted hacking** · May 10th 2009, 05:43 AM

Originally Posted by Pinkk

Well, I'm pretty sure it'll diverge because if I recall correctly from class, since you can express that series as:

$\sum_{k=1}^{\infty}2^{k} -\sum_{k=1}^{\infty}1$

It is pretty obvious $\sum_{k=1}^{\infty}1$ will diverge, so the whole series diverges. Correct me if I'm wrong though.

I made a mistake with the question.

it is supposed to be

$\sum_{k=1}^{\infty}2^{\frac{1}{k}} -1$

Other than that it would be trivial

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