Mixing
Testing series for convergence/divergence
$begingroup$
The task is to test the following series for convergence/ divergence:
$$sum_{n=1}^infty frac{(a+nx)^n}{n!}$$
Now, I have been able to use the Ratio Test and establish that the series converges for $x<1/e$ and diverges for $x>1/e$, but testing the series at $x=1/e$ has been a little more challenging. Could someone tell me how I might get the job done?
sequences-and-series convergence
edited Jan 15 at 19:39
Math1000
19.1k31745
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
$endgroup$
add a comment |
$begingroup$
The task is to test the following series for convergence/ divergence:
$$sum_{n=1}^infty frac{(a+nx)^n}{n!}$$
Now, I have been able to use the Ratio Test and establish that the series converges for $x<1/e$ and diverges for $x>1/e$, but testing the series at $x=1/e$ has been a little more challenging. Could someone tell me how I might get the job done?
sequences-and-series convergence
edited Jan 15 at 19:39
Math1000
19.1k31745
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
$endgroup$
$begingroup$
I think you will need Stirling formula.
$endgroup$
– hamam_Abdallah
Jan 15 at 19:47
add a comment |
$begingroup$
The task is to test the following series for convergence/ divergence:
$$sum_{n=1}^infty frac{(a+nx)^n}{n!}$$
Now, I have been able to use the Ratio Test and establish that the series converges for $x<1/e$ and diverges for $x>1/e$, but testing the series at $x=1/e$ has been a little more challenging. Could someone tell me how I might get the job done?
sequences-and-series convergence
edited Jan 15 at 19:39
Math1000
19.1k31745
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
$endgroup$
The task is to test the following series for convergence/ divergence:
$$sum_{n=1}^infty frac{(a+nx)^n}{n!}$$
Now, I have been able to use the Ratio Test and establish that the series converges for $x<1/e$ and diverges for $x>1/e$, but testing the series at $x=1/e$ has been a little more challenging. Could someone tell me how I might get the job done?
sequences-and-series convergence
sequences-and-series convergence
edited Jan 15 at 19:39
Math1000
19.1k31745
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
edited Jan 15 at 19:39
Math1000
19.1k31745
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
edited Jan 15 at 19:39
Math1000
19.1k31745
edited Jan 15 at 19:39
Math1000
19.1k31745
edited Jan 15 at 19:39
Math1000
19.1k31745
19.1k31745
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
asked Jan 15 at 19:37
s0ulr3aper07s0ulr3aper07
414110
414110
$begingroup$
I think you will need Stirling formula.
$endgroup$
– hamam_Abdallah
Jan 15 at 19:47
add a comment |
$begingroup$
I think you will need Stirling formula.
$endgroup$
– hamam_Abdallah
Jan 15 at 19:47
$begingroup$
I think you will need Stirling formula.
$endgroup$
– hamam_Abdallah
Jan 15 at 19:47
$begingroup$
I think you will need Stirling formula.
$endgroup$
– hamam_Abdallah
Jan 15 at 19:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $x=frac{1}{e}$.
Since $$
frac{(a+frac{n}{e})^n/n!}{(frac{n}{e})^n/n!}=left(1+frac{ae}{n}right)^nto e^{ae},
$$ by the comparison test, $sum_{n=1}^infty a_n<infty $ if and only if $sum_{n=1}^infty frac{(frac{n}{e})^n}{n!}<infty$. So we may assume that $a=0$. By Stirling's formula, we have $$lim_{ntoinfty}frac{n!}{sqrt{2pi n}(frac{n}{e})^n}=1.$$ Since it is a positive sequence with a positive limit, the sequence should be bounded away from $0$ (i.e. have a positive infimum) and have a bounded supremum. So there exist $c>0$ and $C>0$ such that
$$cle frac{sqrt{n}(frac{n}{e})^n}{n!}le C,
$$or equivalently
$$
frac{c}{sqrt{n}}le frac{(frac{n}{e})^n}{n!}le frac{C}{sqrt{n}}.
$$
Since $sum_n frac{1}{sqrt{n}}=infty$, the series diverges for $x=frac{1}{e}$.
If $x=-frac{1}{e}$, then the series becomes alternating eventually. Therefore, the series converges if and only if $|a_n|to 0$ as $nto infty$. And this follows immediately from Stirling's formula:
$$begin{eqnarray}
lim_{ntoinfty}|a_n|&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{sqrt{2pi n}(frac{n}{e})^n}\
&=&lim_{ntoinfty}frac{1}{sqrt{2pi n}}left(1-frac{ae}{n}right)^n=0.
end{eqnarray}$$
There is an alternative approach avoiding use of Stirling's formula. Note that by Taylor series expansion, we have
$$
log(1+t) = t-frac{t^2}{2}+o(t^2).
$$ This implies that there exists $delta>0$ such that
$$
expleft(t-ct^2right)le 1+tle expleft(t-Ct^2right),quadforall 0le tle delta
$$ for some $cin (frac{1}{2},1)$ and $Cin (0,frac{1}{2})$. Let $$a_n = frac{(frac{n}{e})^n}{n!}.$$ Then
$$
frac{a_{n+1}}{a_n}=frac{1}{e}left(1+frac{1}{n}right)^n,
$$ and hence
$$
expleft(-frac{c}{n}right)lefrac{a_{n+1}}{a_n}le expleft(-frac{C}{n}right)
$$ for all sufficiently large $n$. This gives for all large $n$,
$$
kexpleft(-csum_{j=1}^{n-1}frac{1}{j}right)le a_nle Kexpleft(-Csum_{j=1}^{n-1}frac{1}{j}right)
$$ for some $k>0$ and $K>0$. Using the fact that $int_j^{j+1}frac{dt}{t}lefrac{1}{j}=int_{j-1}^jfrac{1}{j}dtleint_{j-1}^jfrac{dt}{t}$ for $jge 2$, we have
$$
log n=int_1^nfrac{dt}{t}lesum_{j=1}^{n-1}frac{1}{j}le 1+int_1^{n-1}frac{dt}{t}le 1+log n.
$$ This in turn implies
$$
ke^{-c}frac{1}{n^c}le a_n le frac{K}{n^C}.
$$ Now, if $x=frac{1}{e}$, then $sum_n a_n =infty$ follows from $a_nge ke^{-c}frac{1}{n^c}$ for all but finitely many $n$. If $x=-frac{1}{e}$, then $frac{|a-frac{n}{e}|^n}{n!}to 0$ follows from
$$begin{eqnarray}
lim_{ntoinfty}frac{|a-frac{n}{e}|^n}{n!}&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&e^{-ae}lim_{ntoinfty}a_n\
&le&e^{-ae}lim_{ntoinfty}frac{K}{n^C}=0.
end{eqnarray}$$
answered Jan 15 at 20:27
SongSong
14.2k1633
$endgroup$
$begingroup$
Could you further elaborate on the part where you introduce c and C?
$endgroup$
– s0ulr3aper07
Jan 15 at 22:06
$begingroup$
@s0ulr3aper07 I hope this makes it clear ...
$endgroup$
– Song
Jan 15 at 22:37
$begingroup$
Re: Stirling's Formula. For many problems like this, it suffices that $L=lim_{nto infty} (n!)^{-1}(n/e)^n sqrt n$ exists and is positive. That $L=1/sqrt {2pi}$ takes a lot more work to prove.
$endgroup$
– DanielWainfleet
Jan 16 at 21:48
add a comment |
$begingroup$
When $x=1/e$ the numerator is equivalent (as $nto infty$) to $(n/e)^n$. Using Stirling's formula for $n!$ your general term is equivalent to $1/sqrt{2pi n}$ and your series is therefore divergent.
answered Jan 15 at 19:59
GReyesGReyes
1,45515
$endgroup$
1
$begingroup$
And what happens if $x=-1/e$? ;-))
$endgroup$
– Mark Viola
Jan 15 at 20:10
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x=frac{1}{e}$.
Since $$
frac{(a+frac{n}{e})^n/n!}{(frac{n}{e})^n/n!}=left(1+frac{ae}{n}right)^nto e^{ae},
$$ by the comparison test, $sum_{n=1}^infty a_n<infty $ if and only if $sum_{n=1}^infty frac{(frac{n}{e})^n}{n!}<infty$. So we may assume that $a=0$. By Stirling's formula, we have $$lim_{ntoinfty}frac{n!}{sqrt{2pi n}(frac{n}{e})^n}=1.$$ Since it is a positive sequence with a positive limit, the sequence should be bounded away from $0$ (i.e. have a positive infimum) and have a bounded supremum. So there exist $c>0$ and $C>0$ such that
$$cle frac{sqrt{n}(frac{n}{e})^n}{n!}le C,
$$or equivalently
$$
frac{c}{sqrt{n}}le frac{(frac{n}{e})^n}{n!}le frac{C}{sqrt{n}}.
$$
Since $sum_n frac{1}{sqrt{n}}=infty$, the series diverges for $x=frac{1}{e}$.
If $x=-frac{1}{e}$, then the series becomes alternating eventually. Therefore, the series converges if and only if $|a_n|to 0$ as $nto infty$. And this follows immediately from Stirling's formula:
$$begin{eqnarray}
lim_{ntoinfty}|a_n|&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{sqrt{2pi n}(frac{n}{e})^n}\
&=&lim_{ntoinfty}frac{1}{sqrt{2pi n}}left(1-frac{ae}{n}right)^n=0.
end{eqnarray}$$
There is an alternative approach avoiding use of Stirling's formula. Note that by Taylor series expansion, we have
$$
log(1+t) = t-frac{t^2}{2}+o(t^2).
$$ This implies that there exists $delta>0$ such that
$$
expleft(t-ct^2right)le 1+tle expleft(t-Ct^2right),quadforall 0le tle delta
$$ for some $cin (frac{1}{2},1)$ and $Cin (0,frac{1}{2})$. Let $$a_n = frac{(frac{n}{e})^n}{n!}.$$ Then
$$
frac{a_{n+1}}{a_n}=frac{1}{e}left(1+frac{1}{n}right)^n,
$$ and hence
$$
expleft(-frac{c}{n}right)lefrac{a_{n+1}}{a_n}le expleft(-frac{C}{n}right)
$$ for all sufficiently large $n$. This gives for all large $n$,
$$
kexpleft(-csum_{j=1}^{n-1}frac{1}{j}right)le a_nle Kexpleft(-Csum_{j=1}^{n-1}frac{1}{j}right)
$$ for some $k>0$ and $K>0$. Using the fact that $int_j^{j+1}frac{dt}{t}lefrac{1}{j}=int_{j-1}^jfrac{1}{j}dtleint_{j-1}^jfrac{dt}{t}$ for $jge 2$, we have
$$
log n=int_1^nfrac{dt}{t}lesum_{j=1}^{n-1}frac{1}{j}le 1+int_1^{n-1}frac{dt}{t}le 1+log n.
$$ This in turn implies
$$
ke^{-c}frac{1}{n^c}le a_n le frac{K}{n^C}.
$$ Now, if $x=frac{1}{e}$, then $sum_n a_n =infty$ follows from $a_nge ke^{-c}frac{1}{n^c}$ for all but finitely many $n$. If $x=-frac{1}{e}$, then $frac{|a-frac{n}{e}|^n}{n!}to 0$ follows from
$$begin{eqnarray}
lim_{ntoinfty}frac{|a-frac{n}{e}|^n}{n!}&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&e^{-ae}lim_{ntoinfty}a_n\
&le&e^{-ae}lim_{ntoinfty}frac{K}{n^C}=0.
end{eqnarray}$$
answered Jan 15 at 20:27
SongSong
14.2k1633
$endgroup$
$begingroup$
Could you further elaborate on the part where you introduce c and C?
$endgroup$
– s0ulr3aper07
Jan 15 at 22:06
$begingroup$
@s0ulr3aper07 I hope this makes it clear ...
$endgroup$
– Song
Jan 15 at 22:37
$begingroup$
Re: Stirling's Formula. For many problems like this, it suffices that $L=lim_{nto infty} (n!)^{-1}(n/e)^n sqrt n$ exists and is positive. That $L=1/sqrt {2pi}$ takes a lot more work to prove.
$endgroup$
– DanielWainfleet
Jan 16 at 21:48
add a comment |
$begingroup$
Let $x=frac{1}{e}$.
Since $$
frac{(a+frac{n}{e})^n/n!}{(frac{n}{e})^n/n!}=left(1+frac{ae}{n}right)^nto e^{ae},
$$ by the comparison test, $sum_{n=1}^infty a_n<infty $ if and only if $sum_{n=1}^infty frac{(frac{n}{e})^n}{n!}<infty$. So we may assume that $a=0$. By Stirling's formula, we have $$lim_{ntoinfty}frac{n!}{sqrt{2pi n}(frac{n}{e})^n}=1.$$ Since it is a positive sequence with a positive limit, the sequence should be bounded away from $0$ (i.e. have a positive infimum) and have a bounded supremum. So there exist $c>0$ and $C>0$ such that
$$cle frac{sqrt{n}(frac{n}{e})^n}{n!}le C,
$$or equivalently
$$
frac{c}{sqrt{n}}le frac{(frac{n}{e})^n}{n!}le frac{C}{sqrt{n}}.
$$
Since $sum_n frac{1}{sqrt{n}}=infty$, the series diverges for $x=frac{1}{e}$.
If $x=-frac{1}{e}$, then the series becomes alternating eventually. Therefore, the series converges if and only if $|a_n|to 0$ as $nto infty$. And this follows immediately from Stirling's formula:
$$begin{eqnarray}
lim_{ntoinfty}|a_n|&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{sqrt{2pi n}(frac{n}{e})^n}\
&=&lim_{ntoinfty}frac{1}{sqrt{2pi n}}left(1-frac{ae}{n}right)^n=0.
end{eqnarray}$$
There is an alternative approach avoiding use of Stirling's formula. Note that by Taylor series expansion, we have
$$
log(1+t) = t-frac{t^2}{2}+o(t^2).
$$ This implies that there exists $delta>0$ such that
$$
expleft(t-ct^2right)le 1+tle expleft(t-Ct^2right),quadforall 0le tle delta
$$ for some $cin (frac{1}{2},1)$ and $Cin (0,frac{1}{2})$. Let $$a_n = frac{(frac{n}{e})^n}{n!}.$$ Then
$$
frac{a_{n+1}}{a_n}=frac{1}{e}left(1+frac{1}{n}right)^n,
$$ and hence
$$
expleft(-frac{c}{n}right)lefrac{a_{n+1}}{a_n}le expleft(-frac{C}{n}right)
$$ for all sufficiently large $n$. This gives for all large $n$,
$$
kexpleft(-csum_{j=1}^{n-1}frac{1}{j}right)le a_nle Kexpleft(-Csum_{j=1}^{n-1}frac{1}{j}right)
$$ for some $k>0$ and $K>0$. Using the fact that $int_j^{j+1}frac{dt}{t}lefrac{1}{j}=int_{j-1}^jfrac{1}{j}dtleint_{j-1}^jfrac{dt}{t}$ for $jge 2$, we have
$$
log n=int_1^nfrac{dt}{t}lesum_{j=1}^{n-1}frac{1}{j}le 1+int_1^{n-1}frac{dt}{t}le 1+log n.
$$ This in turn implies
$$
ke^{-c}frac{1}{n^c}le a_n le frac{K}{n^C}.
$$ Now, if $x=frac{1}{e}$, then $sum_n a_n =infty$ follows from $a_nge ke^{-c}frac{1}{n^c}$ for all but finitely many $n$. If $x=-frac{1}{e}$, then $frac{|a-frac{n}{e}|^n}{n!}to 0$ follows from
$$begin{eqnarray}
lim_{ntoinfty}frac{|a-frac{n}{e}|^n}{n!}&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&e^{-ae}lim_{ntoinfty}a_n\
&le&e^{-ae}lim_{ntoinfty}frac{K}{n^C}=0.
end{eqnarray}$$
answered Jan 15 at 20:27
SongSong
14.2k1633
$endgroup$
$begingroup$
Could you further elaborate on the part where you introduce c and C?
$endgroup$
– s0ulr3aper07
Jan 15 at 22:06
$begingroup$
@s0ulr3aper07 I hope this makes it clear ...
$endgroup$
– Song
Jan 15 at 22:37
$begingroup$
Re: Stirling's Formula. For many problems like this, it suffices that $L=lim_{nto infty} (n!)^{-1}(n/e)^n sqrt n$ exists and is positive. That $L=1/sqrt {2pi}$ takes a lot more work to prove.
$endgroup$
– DanielWainfleet
Jan 16 at 21:48
add a comment |
$begingroup$
Let $x=frac{1}{e}$.
Since $$
frac{(a+frac{n}{e})^n/n!}{(frac{n}{e})^n/n!}=left(1+frac{ae}{n}right)^nto e^{ae},
$$ by the comparison test, $sum_{n=1}^infty a_n<infty $ if and only if $sum_{n=1}^infty frac{(frac{n}{e})^n}{n!}<infty$. So we may assume that $a=0$. By Stirling's formula, we have $$lim_{ntoinfty}frac{n!}{sqrt{2pi n}(frac{n}{e})^n}=1.$$ Since it is a positive sequence with a positive limit, the sequence should be bounded away from $0$ (i.e. have a positive infimum) and have a bounded supremum. So there exist $c>0$ and $C>0$ such that
$$cle frac{sqrt{n}(frac{n}{e})^n}{n!}le C,
$$or equivalently
$$
frac{c}{sqrt{n}}le frac{(frac{n}{e})^n}{n!}le frac{C}{sqrt{n}}.
$$
Since $sum_n frac{1}{sqrt{n}}=infty$, the series diverges for $x=frac{1}{e}$.
If $x=-frac{1}{e}$, then the series becomes alternating eventually. Therefore, the series converges if and only if $|a_n|to 0$ as $nto infty$. And this follows immediately from Stirling's formula:
$$begin{eqnarray}
lim_{ntoinfty}|a_n|&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{sqrt{2pi n}(frac{n}{e})^n}\
&=&lim_{ntoinfty}frac{1}{sqrt{2pi n}}left(1-frac{ae}{n}right)^n=0.
end{eqnarray}$$
There is an alternative approach avoiding use of Stirling's formula. Note that by Taylor series expansion, we have
$$
log(1+t) = t-frac{t^2}{2}+o(t^2).
$$ This implies that there exists $delta>0$ such that
$$
expleft(t-ct^2right)le 1+tle expleft(t-Ct^2right),quadforall 0le tle delta
$$ for some $cin (frac{1}{2},1)$ and $Cin (0,frac{1}{2})$. Let $$a_n = frac{(frac{n}{e})^n}{n!}.$$ Then
$$
frac{a_{n+1}}{a_n}=frac{1}{e}left(1+frac{1}{n}right)^n,
$$ and hence
$$
expleft(-frac{c}{n}right)lefrac{a_{n+1}}{a_n}le expleft(-frac{C}{n}right)
$$ for all sufficiently large $n$. This gives for all large $n$,
$$
kexpleft(-csum_{j=1}^{n-1}frac{1}{j}right)le a_nle Kexpleft(-Csum_{j=1}^{n-1}frac{1}{j}right)
$$ for some $k>0$ and $K>0$. Using the fact that $int_j^{j+1}frac{dt}{t}lefrac{1}{j}=int_{j-1}^jfrac{1}{j}dtleint_{j-1}^jfrac{dt}{t}$ for $jge 2$, we have
$$
log n=int_1^nfrac{dt}{t}lesum_{j=1}^{n-1}frac{1}{j}le 1+int_1^{n-1}frac{dt}{t}le 1+log n.
$$ This in turn implies
$$
ke^{-c}frac{1}{n^c}le a_n le frac{K}{n^C}.
$$ Now, if $x=frac{1}{e}$, then $sum_n a_n =infty$ follows from $a_nge ke^{-c}frac{1}{n^c}$ for all but finitely many $n$. If $x=-frac{1}{e}$, then $frac{|a-frac{n}{e}|^n}{n!}to 0$ follows from
$$begin{eqnarray}
lim_{ntoinfty}frac{|a-frac{n}{e}|^n}{n!}&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&e^{-ae}lim_{ntoinfty}a_n\
&le&e^{-ae}lim_{ntoinfty}frac{K}{n^C}=0.
end{eqnarray}$$
answered Jan 15 at 20:27
SongSong
14.2k1633
$endgroup$
Let $x=frac{1}{e}$.
Since $$
frac{(a+frac{n}{e})^n/n!}{(frac{n}{e})^n/n!}=left(1+frac{ae}{n}right)^nto e^{ae},
$$ by the comparison test, $sum_{n=1}^infty a_n<infty $ if and only if $sum_{n=1}^infty frac{(frac{n}{e})^n}{n!}<infty$. So we may assume that $a=0$. By Stirling's formula, we have $$lim_{ntoinfty}frac{n!}{sqrt{2pi n}(frac{n}{e})^n}=1.$$ Since it is a positive sequence with a positive limit, the sequence should be bounded away from $0$ (i.e. have a positive infimum) and have a bounded supremum. So there exist $c>0$ and $C>0$ such that
$$cle frac{sqrt{n}(frac{n}{e})^n}{n!}le C,
$$or equivalently
$$
frac{c}{sqrt{n}}le frac{(frac{n}{e})^n}{n!}le frac{C}{sqrt{n}}.
$$
Since $sum_n frac{1}{sqrt{n}}=infty$, the series diverges for $x=frac{1}{e}$.
If $x=-frac{1}{e}$, then the series becomes alternating eventually. Therefore, the series converges if and only if $|a_n|to 0$ as $nto infty$. And this follows immediately from Stirling's formula:
$$begin{eqnarray}
lim_{ntoinfty}|a_n|&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{sqrt{2pi n}(frac{n}{e})^n}\
&=&lim_{ntoinfty}frac{1}{sqrt{2pi n}}left(1-frac{ae}{n}right)^n=0.
end{eqnarray}$$
There is an alternative approach avoiding use of Stirling's formula. Note that by Taylor series expansion, we have
$$
log(1+t) = t-frac{t^2}{2}+o(t^2).
$$ This implies that there exists $delta>0$ such that
$$
expleft(t-ct^2right)le 1+tle expleft(t-Ct^2right),quadforall 0le tle delta
$$ for some $cin (frac{1}{2},1)$ and $Cin (0,frac{1}{2})$. Let $$a_n = frac{(frac{n}{e})^n}{n!}.$$ Then
$$
frac{a_{n+1}}{a_n}=frac{1}{e}left(1+frac{1}{n}right)^n,
$$ and hence
$$
expleft(-frac{c}{n}right)lefrac{a_{n+1}}{a_n}le expleft(-frac{C}{n}right)
$$ for all sufficiently large $n$. This gives for all large $n$,
$$
kexpleft(-csum_{j=1}^{n-1}frac{1}{j}right)le a_nle Kexpleft(-Csum_{j=1}^{n-1}frac{1}{j}right)
$$ for some $k>0$ and $K>0$. Using the fact that $int_j^{j+1}frac{dt}{t}lefrac{1}{j}=int_{j-1}^jfrac{1}{j}dtleint_{j-1}^jfrac{dt}{t}$ for $jge 2$, we have
$$
log n=int_1^nfrac{dt}{t}lesum_{j=1}^{n-1}frac{1}{j}le 1+int_1^{n-1}frac{dt}{t}le 1+log n.
$$ This in turn implies
$$
ke^{-c}frac{1}{n^c}le a_n le frac{K}{n^C}.
$$ Now, if $x=frac{1}{e}$, then $sum_n a_n =infty$ follows from $a_nge ke^{-c}frac{1}{n^c}$ for all but finitely many $n$. If $x=-frac{1}{e}$, then $frac{|a-frac{n}{e}|^n}{n!}to 0$ follows from
$$begin{eqnarray}
lim_{ntoinfty}frac{|a-frac{n}{e}|^n}{n!}&=& lim_{ntoinfty}frac{(frac{n}{e}-a)^n}{n!}\
&=&e^{-ae}lim_{ntoinfty}a_n\
&le&e^{-ae}lim_{ntoinfty}frac{K}{n^C}=0.
end{eqnarray}$$
answered Jan 15 at 20:27
SongSong
14.2k1633
edited Jan 16 at 10:31
answered Jan 15 at 20:27
SongSong
14.2k1633
answered Jan 15 at 20:27
SongSong
14.2k1633
answered Jan 15 at 20:27
SongSong
14.2k1633
14.2k1633
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Could you further elaborate on the part where you introduce c and C?
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– s0ulr3aper07
Jan 15 at 22:06
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@s0ulr3aper07 I hope this makes it clear ...
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– Song
Jan 15 at 22:37
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Re: Stirling's Formula. For many problems like this, it suffices that $L=lim_{nto infty} (n!)^{-1}(n/e)^n sqrt n$ exists and is positive. That $L=1/sqrt {2pi}$ takes a lot more work to prove.
$endgroup$
– DanielWainfleet
Jan 16 at 21:48
add a comment |
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Could you further elaborate on the part where you introduce c and C?
$endgroup$
– s0ulr3aper07
Jan 15 at 22:06
$begingroup$
@s0ulr3aper07 I hope this makes it clear ...
$endgroup$
– Song
Jan 15 at 22:37
$begingroup$
Re: Stirling's Formula. For many problems like this, it suffices that $L=lim_{nto infty} (n!)^{-1}(n/e)^n sqrt n$ exists and is positive. That $L=1/sqrt {2pi}$ takes a lot more work to prove.
$endgroup$
– DanielWainfleet
Jan 16 at 21:48
$begingroup$
Could you further elaborate on the part where you introduce c and C?
$endgroup$
– s0ulr3aper07
Jan 15 at 22:06
$begingroup$
Could you further elaborate on the part where you introduce c and C?
$endgroup$
– s0ulr3aper07
Jan 15 at 22:06
$begingroup$
@s0ulr3aper07 I hope this makes it clear ...
$endgroup$
– Song
Jan 15 at 22:37
$begingroup$
@s0ulr3aper07 I hope this makes it clear ...
$endgroup$
– Song
Jan 15 at 22:37
$begingroup$
Re: Stirling's Formula. For many problems like this, it suffices that $L=lim_{nto infty} (n!)^{-1}(n/e)^n sqrt n$ exists and is positive. That $L=1/sqrt {2pi}$ takes a lot more work to prove.
$endgroup$
– DanielWainfleet
Jan 16 at 21:48
$begingroup$
Re: Stirling's Formula. For many problems like this, it suffices that $L=lim_{nto infty} (n!)^{-1}(n/e)^n sqrt n$ exists and is positive. That $L=1/sqrt {2pi}$ takes a lot more work to prove.
$endgroup$
– DanielWainfleet
Jan 16 at 21:48
add a comment |
$begingroup$
When $x=1/e$ the numerator is equivalent (as $nto infty$) to $(n/e)^n$. Using Stirling's formula for $n!$ your general term is equivalent to $1/sqrt{2pi n}$ and your series is therefore divergent.
answered Jan 15 at 19:59
GReyesGReyes
1,45515
$endgroup$
1
$begingroup$
And what happens if $x=-1/e$? ;-))
$endgroup$
– Mark Viola
Jan 15 at 20:10
add a comment |
$begingroup$
When $x=1/e$ the numerator is equivalent (as $nto infty$) to $(n/e)^n$. Using Stirling's formula for $n!$ your general term is equivalent to $1/sqrt{2pi n}$ and your series is therefore divergent.
answered Jan 15 at 19:59
GReyesGReyes
1,45515
$endgroup$
1
$begingroup$
And what happens if $x=-1/e$? ;-))
$endgroup$
– Mark Viola
Jan 15 at 20:10
add a comment |
$begingroup$
When $x=1/e$ the numerator is equivalent (as $nto infty$) to $(n/e)^n$. Using Stirling's formula for $n!$ your general term is equivalent to $1/sqrt{2pi n}$ and your series is therefore divergent.
answered Jan 15 at 19:59
GReyesGReyes
1,45515
$endgroup$
When $x=1/e$ the numerator is equivalent (as $nto infty$) to $(n/e)^n$. Using Stirling's formula for $n!$ your general term is equivalent to $1/sqrt{2pi n}$ and your series is therefore divergent.
answered Jan 15 at 19:59
GReyesGReyes
1,45515
answered Jan 15 at 19:59
GReyesGReyes
1,45515
answered Jan 15 at 19:59
GReyesGReyes
1,45515
answered Jan 15 at 19:59
GReyesGReyes
1,45515
1,45515
1
$begingroup$
And what happens if $x=-1/e$? ;-))
$endgroup$
– Mark Viola
Jan 15 at 20:10
add a comment |
1
$begingroup$
And what happens if $x=-1/e$? ;-))
$endgroup$
– Mark Viola
Jan 15 at 20:10
1
1
$begingroup$
And what happens if $x=-1/e$? ;-))
$endgroup$
– Mark Viola
Jan 15 at 20:10
$begingroup$
And what happens if $x=-1/e$? ;-))
$endgroup$
– Mark Viola
Jan 15 at 20:10
add a comment |
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I think you will need Stirling formula.
$endgroup$
– hamam_Abdallah
Jan 15 at 19:47