Math Help - Help with series

1. Help with series

Determine if the following are absolutely convergent, conditionally convergent or divergent.

(1) $\sum_{n=0}^{\infty }{\frac { 1!+2!+...+n!}{ \left(2n \right)!}}$

I am really quite confused as to how to work this out. I would be grateful if someone could point me in the right direction by telling me where I've gone wrong.

I've used the ratio test and I have a limit that still has factorials in it. But I don't know any other test that would be able to help me get rid of the factorials.

2. first whit testo comparation

$
\sum\limits_{n = 1}^\infty {\frac{{1! + 2! + 3! + ... + n!}}
{{\left( {2n} \right)!}}} \leqslant \sum\limits_{n = 1}^\infty {\frac{{n! + n! + n! + ... + n!}}
{{\left( {2n} \right)!}}} = \sum\limits_{n = 1}^\infty {\frac{{n \cdot n!}}
{{\left( {2n} \right)!}}}
$

If $
a_n = \frac{{n \cdot n!}}
{{\left( {2n} \right)!}}
$
now I use the quotien test

$
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}
{{a_n }}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right) \cdot \left( {n + 1} \right)!}}
{{\left( {2(n + 1)} \right)!}} \cdot \frac{{\left( {2n} \right)!}}
{{n \cdot n!}} =
$
$
\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)\left( {n + 1} \right) \cdot n!}}
{{\left( {2n + 2} \right)\left( {2n + 1} \right)\left( {2n} \right)!}} \cdot \frac{{\left( {2n} \right)!}}
{{n \cdot n!}}
$

$
= \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)\left( {n + 1} \right)}}
{{\left( {2n + 1} \right)\left( {2n + 1} \right)}} \cdot \frac{1}
{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {1 + \frac{1}
{n}} \right)\left( {1 + \frac{1}
{n}} \right) \cdot \frac{1}
{n}}}
{{\left( {2 + \frac{2}
{n}} \right)\left( {2 + \frac{1}
{n}} \right)}}
$
$
= \frac{{\left( {1 + 0} \right)^2 \cdot 0}}
{{\left( {2 + 0} \right)^2 }} = 0 < 1
$

So, how $
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}
{{a_n }}} \right| < 1 \Rightarrow
$
the serie converge

I forget say you that if this serie converge, then (1) converge too. And how converge whit and whitout absolute value, the serie is absolutely convergent

3. Originally Posted by Hweengee
Determine if the following are absolutely convergent, conditionally convergent or divergent.

(1) $\sum_{n=0}^{\infty }{\frac { 1!+2!+...+n!}{ \left(2n \right)!}}$

I am really quite confused as to how to work this out. I would be grateful if someone could point me in the right direction by telling me where I've gone wrong.

I've used the ratio test and I have a limit that still has factorials in it. But I don't know any other test that would be able to help me get rid of the factorials.
There isn't even a need to make a comparison

$\lim_{n\to\infty}\left(\frac{1!+2!+\cdots{+n!}}{(2 n)!}\right)^{\frac{1}{n}}\sim\lim_{n\to\infty}\lef t(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}=0$

Therefore the serise converges

4. What's fun with you Mathstud is that everything is obvious, clear or simple and everybody knows the limits you know

I don't doubt your limit $\lim_{n\to\infty}\left(\frac{1}{(2n)!}\right)^{\fr ac{1}{n}}=0$ is correct, however, I've been taught $0^0$ is not defined (by convention, it's 1), so it is not that obvious.

A brief reminder of Stirling's approximation formula would be nice.

Tsssk, try to imagine that some people here are not as skilled as you are, please.

5. Originally Posted by Moo
What's fun with you Mathstud is that everything is obvious, clear or simple and everybody knows the limits you know

I don't doubt your limit $\lim_{n\to\infty}\left(\frac{1}{(2n)!}\right)^{\fr ac{1}{n}}=0$ is correct, however, I've been taught $0^0$ is not defined (by convention, it's 1), so it is not that obvious.

A brief reminder of Stirling's approximation formula would be nice.

Tsssk, try to imagine that some people here are not as skilled as you are, please.
I am really sorry. I will try to be more dilligent. I can supply the proof if anyone would like it is really interesting but we have that

$n!\sim\sqrt{2\pi{n}}n^ne^{-n}$

What this means is that as n approaches infinity the functions become equivalent.

From there we can direct substitute to get

$(an)!\sim\sqrt{2a\pi{n}}(an)^{an}e^{-an}=\sqrt{2a\pi{n}}a^{an}\left(n^n\right)^a(e^{-n})^a$

So we can see that

$(2n)!\sim\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2$

So

$\lim_{n\to\infty}\left(\frac{n!}{(2n)!}\right)^{\f rac{1}{n}}\sim\lim_{n\to\infty}\left(\frac{\sqrt{2 \pi{n}}n^ne^{-n}}{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^2}\right)^{\frac{1}{n}}$

$=\lim_{n\to\infty}\left(\frac{1}{\sqrt{2}n^ne^{-n}}\right)^{\frac{1}{n}}$

$=\lim_{n\to\infty}\frac{e}{2^{\frac{1}{2n}}n}\to{0 }$

And since

$0\leq\frac{1}{(2n)!}\leq\frac{n!}{(2n)!}$

It can be seen pretty easily by the squeeze theorem that the limit of the middle term is zero as well

I hope that suffices.

6. This is an interesting factoring exercise.
$\left( {2n} \right)! = 2^n \left( {n!} \right)(2n - 1)(2n - 3) \cdots (3)(1)$

$\frac{{\sum\limits_{k = 1}^n {k!} }}{{\left( {2n} \right)!}} \leqslant \frac{n}{{2^n }}$

7. haha the thing is ~ denotes asymptotic equivalence? something which i have not even learnt. thanks anyway.

8. Originally Posted by Mathstud28
I am really sorry. I will try to be more dilligent. I can supply the proof if anyone would like it is really interesting but we have that

$n!\sim\sqrt{2\pi{n}}n^ne^{-n}$

What this means is that as n approaches infinity the functions become equivalent.

From there we can direct substitute to get

$(an)!\sim\sqrt{2a\pi{n}}(an)^{an}e^{-an}=\sqrt{2a\pi{n}}a^{an}\left(n^n\right)^a(e^{-n})^a$

So we can see that

$(2n)!\sim\sqrt{4\pi{n}}4^n(n^n)^2(e^{-n})^2$

So

$\lim_{n\to\infty}\left(\frac{n!}{(2n)!}\right)^{\f rac{1}{n}}\sim\lim_{n\to\infty}\left(\frac{\sqrt{2 \pi{n}}n^ne^{-n}}{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^2}\right)^{\frac{1}{n}}$

$=\lim_{n\to\infty}\left(\frac{1}{\sqrt{2}n^ne^{-n}}\right)^{\frac{1}{n}}$

$=\lim_{n\to\infty}\frac{e}{2^{\frac{1}{2n}}n}\to{0 }$

And since

$0\leq\frac{1}{(2n)!}\leq\frac{n!}{(2n)!}$

It can be seen pretty easily by the squeeze theorem that the limit of the middle term is zero as well

I hope that suffices.
Small problem:

$\left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}$

is not the $n$-th root of the $n$-th term of the series, so there is still a gap to fill in your argument.

You need to show that considering the above is equivalent to considering:

$\left[ \frac { 1!+2!+...+n!}{ \left(2n \right)!} \right] ^{1/n}$

CB

9. Originally Posted by CaptainBlack
Small problem:

$\left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}$

is not the $n$-th root of the $n$-th term of the series, so there is still a gap to fill in your argument.

You need to show that considering the above is equivalent to considering:

$\left[ \frac { 1!+2!+...+n!}{ \left(2n \right)!} \right] ^{1/n}$

CB
Yes but they are asymptotically equivalent. I will come back tonight and show explicitly, but if you are looking for proof of that taking the logaritim of the limit makes this more obvious

Or just see

$\lim_{n\to\infty}\left(\frac{1!+2!+\cdots+n!}{(2n) !}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\left(n!\ cdot\frac{\frac{1}{n!}+\frac{2!}{n!}+\cdots+1}{(2n )!}\right)^{\frac{1}{n}}$

$=\lim_{n\to\infty}\left(\frac{1}{n!}+\frac{2!}{n!} +\cdots+1\right)^{\frac{1}{n}}\cdot\lim_{n\to\inft y}\left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}=1\cd ot{0}=0$

10. Originally Posted by Mathstud28
Yes but they are asymptotically equivalent. I will come back tonight and show explicitly, but if you are looking for proof of that taking the logaritim of the limit makes this more obvious

Or just see

$\lim_{n\to\infty}\left(\frac{1!+2!+\cdots+n!}{(2n) !}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\left(n!\ cdot\frac{\frac{1}{n!}+\frac{2!}{n!}+\cdots+1}{(2n )!}\right)^{\frac{1}{n}}$

$=\lim_{n\to\infty}\left(\frac{1}{n!}+\frac{2!}{n!} +\cdots+1\right)^{\frac{1}{n}}\cdot\lim_{n\to\inft y}\left(\frac{n!}{(2n)!}\right)^{\frac{1}{n}}=1\cd ot{0}=0$
I didn't say it could not be done, just that you have to do it. but it leaves this as a very clumsy method when you put in the detail you have tried to wave your hands over.

After all hand waving is a pointless activity with this series, since it is obviously convergent we could just have made that claim if we were going to handwave about it.

CB

11. Originally Posted by CaptainBlack
I didn't say it could not be done, just that you have to do it. but it leaves this as a very clumsy method when you put in the detail you have tried to wave your hands over.

After all hand waving is a pointless activity with this series, since it is obviously convergent we could just have made that claim if we were going to handwave about it.

CB
Sorry if I did something improper, In all honesty it isn't that laborious in my opinion. As for the hand-waving I totally understand where your coming from but for replying on a math forum it gives a solution but without an integral part, left for the poster. This helps them learn something. And I was just giving an alternate solution. I don't know, that's just my opinion. Thank you for your input.