# Divergence or Convergence of Series

• Jan 26th 2011, 04:20 AM
4Math
Divergence or Convergence of Series
Hi,

I have have to determine if the following series is convergent or divergent.

$\displaystyle \sum_{k= 1}^{\infty} \frac{(k!)^{\frac{1}{k}}}{k}}= \sum_{k= 1}^{\infty} \root k \of {k!}{\frac{1}{k}$

$\displaystyle a_k=\frac{(k!)^{\frac{1}{k}}}{k}}$

$\displaystyle b_k=\frac{1}{k}}}$

$b_k$ is a p-series and diverges.

$\displaystyle\lim_{k \to \infty} \frac{a_k}{b_k}}=\root k \of {k!}=1$

According to direct comparison rule, since $b_k$ is divergent then the given series must also be divergent.(1)

Is (1)correct?

I am also unsure about the following limit:

(2) $\displaystyle\lim_{k \to \infty} \sqrt{(k!)}=1$

Is (2) correct?

I would appreciate a response.

Thank you
• Jan 26th 2011, 04:35 AM
CaptainBlack
Quote:

Originally Posted by 4Math
Hi,

I have have to determine if the following series is convergent or divergent.

$\displaystyle \sum_{k= 1}^{\infty} \frac{(k!)^{\frac{1}{k}}}{k}}= \sum_{k= 1}^{\infty} \root k \of {k!}{\frac{1}{k}$

$\displaystyle a_k=\frac{(k!)^{\frac{1}{k}}}{k}}$

$\displaystyle b_k=\frac{1}{k}}}$

$b_k$ is a p-series and diverges.

$\displaystyle\lim_{k \to \infty} \frac{a_k}{b_k}}=\root k \of {k!}=1$

According to direct comparison rule, since $b_k$ is divergent then the given series must also be divergent.(1)

Is (1)correct?

I am also unsure about the following limit:

(2) $\displaystyle\lim_{k \to \infty} \sqrt{(k!)}=1$

Is (2) correct?

I would appreciate a response.

Thank you

Why do you think that:

$\displaystyle \lim_{n\to \infty} (n!)^{1/n}=1$

What light does Stirlings formula throw on this limit?

CB
• Jan 26th 2011, 06:54 AM
4Math
Quote:

Originally Posted by CaptainBlack
Why do you think that:

$\displaystyle \lim_{n\to \infty} (n!)^{1/n}=1$

What light does Stirlings formula throw on this limit?

CB

The correct limit should be:

$\displaystyle\lim_{k \to \infty} \sqrt{(k!)}\rightarrow \infty$ when $k \to \infty$

Unfortunately I cannot see how Stilrlings formula can be applied and/or relate to this limit. Can you please show this.

Kind regards
• Jan 26th 2011, 08:18 AM
FernandoRevilla
Quote:

Originally Posted by 4Math
Unfortunately I cannot see how Stilrlings formula can be applied and/or relate to this limit. Can you please show this.

Using Stirling's formula:

$\displaystyle\lim_{k \to{+}\infty}{\dfrac{\sqrt[k]{k!}}{k}}=\displaystyle\lim_{k \to{+}\infty}\dfrac{(2\pi k)^{\frac{1}{2k}}ke^{-1}}{k}=\ldots=\dfrac{1}{e}\neq 0$

So, the series is divergent.

Fernando Revilla
• Jan 26th 2011, 10:40 AM
4Math
Quote:

Originally Posted by FernandoRevilla
Using Stirling's formula:

$\displaystyle\lim_{k \to{+}\infty}{\dfrac{\sqrt[k]{k!}}{k}}=\displaystyle\lim_{k \to{+}\infty}\dfrac{(2\pi k)^{\frac{1}{2k}}ke^{-1}}{k}=\ldots=\dfrac{1}{e}\neq 0$

So, the series is divergent.

Fernando Revilla

Can I also use Direct Comparison Test like I have on my first post and if so have I done it correctly?

Kind regards
• Jan 26th 2011, 10:54 AM
FernandoRevilla
Quote:

Originally Posted by 4Math
Can I also use Direct Comparison Test like I have on my first post and if so have I done it correctly?

Yes, you can. But,

$\displaystyle\lim_{k \to{+}\infty}{\dfrac{a_n}{b_n}}=+\infty$

and

$\displaystyle\sum_{k=1}^{+\infty}b_n$

is divergent, so the given series is divergent.

Fernando Revilla