Limit of terms of Harmonic series
$begingroup$
Let $s(n)$ be the smallest positive integer such that $1+frac{1}{2}+frac{1}{3}+...+1/s(n)geq{n}$
Find $displaystylelim_{ntoinfty} frac{s(n+1)}{s(n)}$
The values of $s(n)$ are
$s(1)=1, s(2)=4, s(3)=11, s(4)=31, s(5)=83, s(6)=227, s(7)=616 $ and so on
$frac{s(2)}{s(1)}=4, frac{s(3)}{s(2)}=2.75, frac{s(4)}{s(3)}=2.8181..., frac{s(5)}{s(4)}=2.677..., frac{s(6)}{s(5)}=2.73493..., frac{s(7)}{s(6)}=2.713...,$
I currently can't find a formula for $s(n)$ but from the calculations I can see that the ratio converges to $e$.
Any tips on how to prove this?
sequences-and-series
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
$endgroup$
add a comment |
$begingroup$
Let $s(n)$ be the smallest positive integer such that $1+frac{1}{2}+frac{1}{3}+...+1/s(n)geq{n}$
Find $displaystylelim_{ntoinfty} frac{s(n+1)}{s(n)}$
The values of $s(n)$ are
$s(1)=1, s(2)=4, s(3)=11, s(4)=31, s(5)=83, s(6)=227, s(7)=616 $ and so on
$frac{s(2)}{s(1)}=4, frac{s(3)}{s(2)}=2.75, frac{s(4)}{s(3)}=2.8181..., frac{s(5)}{s(4)}=2.677..., frac{s(6)}{s(5)}=2.73493..., frac{s(7)}{s(6)}=2.713...,$
I currently can't find a formula for $s(n)$ but from the calculations I can see that the ratio converges to $e$.
Any tips on how to prove this?
sequences-and-series
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
$endgroup$
add a comment |
$begingroup$
Let $s(n)$ be the smallest positive integer such that $1+frac{1}{2}+frac{1}{3}+...+1/s(n)geq{n}$
Find $displaystylelim_{ntoinfty} frac{s(n+1)}{s(n)}$
The values of $s(n)$ are
$s(1)=1, s(2)=4, s(3)=11, s(4)=31, s(5)=83, s(6)=227, s(7)=616 $ and so on
$frac{s(2)}{s(1)}=4, frac{s(3)}{s(2)}=2.75, frac{s(4)}{s(3)}=2.8181..., frac{s(5)}{s(4)}=2.677..., frac{s(6)}{s(5)}=2.73493..., frac{s(7)}{s(6)}=2.713...,$
I currently can't find a formula for $s(n)$ but from the calculations I can see that the ratio converges to $e$.
Any tips on how to prove this?
sequences-and-series
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
$endgroup$
Let $s(n)$ be the smallest positive integer such that $1+frac{1}{2}+frac{1}{3}+...+1/s(n)geq{n}$
Find $displaystylelim_{ntoinfty} frac{s(n+1)}{s(n)}$
The values of $s(n)$ are
$s(1)=1, s(2)=4, s(3)=11, s(4)=31, s(5)=83, s(6)=227, s(7)=616 $ and so on
$frac{s(2)}{s(1)}=4, frac{s(3)}{s(2)}=2.75, frac{s(4)}{s(3)}=2.8181..., frac{s(5)}{s(4)}=2.677..., frac{s(6)}{s(5)}=2.73493..., frac{s(7)}{s(6)}=2.713...,$
I currently can't find a formula for $s(n)$ but from the calculations I can see that the ratio converges to $e$.
Any tips on how to prove this?
sequences-and-series
sequences-and-series
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
asked Dec 6 '18 at 14:16
A. BrunoA. Bruno
1035
1035
add a comment |
add a comment |
1 Answer
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$begingroup$
The point is that the partial sums of the harmonic series are slowly increasing:
$$ H_n = sum_{k=1}^{n}frac{1}{k} = log n + underbrace{sum_{kgeq 1}left[frac{1}{n}-logleft(1+frac{1}{n}right)right]}_{gamma,approxfrac{1}{sqrt{3}}}+Oleft(frac{1}{n}right) $$
hence for any $varepsilon>0$ and for any $N$ large enough, the smallest $n$ such that $H_ngeq N$ is bounded between $expleft(N-gamma-varepsilonright)$ and $exp(N-gamma+varepsilon)$. In particular the wanted limit equals $exp(1)=e$.
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The point is that the partial sums of the harmonic series are slowly increasing:
$$ H_n = sum_{k=1}^{n}frac{1}{k} = log n + underbrace{sum_{kgeq 1}left[frac{1}{n}-logleft(1+frac{1}{n}right)right]}_{gamma,approxfrac{1}{sqrt{3}}}+Oleft(frac{1}{n}right) $$
hence for any $varepsilon>0$ and for any $N$ large enough, the smallest $n$ such that $H_ngeq N$ is bounded between $expleft(N-gamma-varepsilonright)$ and $exp(N-gamma+varepsilon)$. In particular the wanted limit equals $exp(1)=e$.
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
The point is that the partial sums of the harmonic series are slowly increasing:
$$ H_n = sum_{k=1}^{n}frac{1}{k} = log n + underbrace{sum_{kgeq 1}left[frac{1}{n}-logleft(1+frac{1}{n}right)right]}_{gamma,approxfrac{1}{sqrt{3}}}+Oleft(frac{1}{n}right) $$
hence for any $varepsilon>0$ and for any $N$ large enough, the smallest $n$ such that $H_ngeq N$ is bounded between $expleft(N-gamma-varepsilonright)$ and $exp(N-gamma+varepsilon)$. In particular the wanted limit equals $exp(1)=e$.
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
The point is that the partial sums of the harmonic series are slowly increasing:
$$ H_n = sum_{k=1}^{n}frac{1}{k} = log n + underbrace{sum_{kgeq 1}left[frac{1}{n}-logleft(1+frac{1}{n}right)right]}_{gamma,approxfrac{1}{sqrt{3}}}+Oleft(frac{1}{n}right) $$
hence for any $varepsilon>0$ and for any $N$ large enough, the smallest $n$ such that $H_ngeq N$ is bounded between $expleft(N-gamma-varepsilonright)$ and $exp(N-gamma+varepsilon)$. In particular the wanted limit equals $exp(1)=e$.
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
The point is that the partial sums of the harmonic series are slowly increasing:
$$ H_n = sum_{k=1}^{n}frac{1}{k} = log n + underbrace{sum_{kgeq 1}left[frac{1}{n}-logleft(1+frac{1}{n}right)right]}_{gamma,approxfrac{1}{sqrt{3}}}+Oleft(frac{1}{n}right) $$
hence for any $varepsilon>0$ and for any $N$ large enough, the smallest $n$ such that $H_ngeq N$ is bounded between $expleft(N-gamma-varepsilonright)$ and $exp(N-gamma+varepsilon)$. In particular the wanted limit equals $exp(1)=e$.
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 6 '18 at 14:31
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
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