Let $sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}$, does it converge or does it diverge?
Let $displaystyle sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}$, does it converge or does it diverge?
I'm not completely sure that my calculation is correct, check it please.
$$begin{align}sum_{nge 0} frac{(-1)^{n+1}}{3 n+6 (-1)^n}&=-frac13sum_{nge 0} frac{(-1)^{n}}{n+2 (-1)^n}cdotfrac{(-1)^n}{(-1)^n}\&=-frac13sum_{nge 0} frac1{n(-1)^n+2}\&=-frac13lim_{mtoinfty}left(sum_{n=0\ 2mid n}^mfrac1{n+2}+sum_{n=0\ 2nmid n}^mfrac1{-n+2}right)\&=-frac13lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)end{align}$$
If $m$ is even then
$$begin{align}lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)&=lim_{mtoinfty}left(left(sum_{n=1}^{m-2}frac{(-1)^{n}}nright)+frac1m+frac1{m+2}+1right)\&=-log (2)+1end{align}$$
If $m$ is odd then
$$lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)=lim_{mtoinfty}left(left(sum_{n=1}^{m-1}frac{(-1)^{n}}nright)+frac1{m+1}+1right)=-log (2)+1$$
Then finally
$$bbox[2pt,border:yellow solid 2px]{sum_{n=0}^inftyfrac{(-1)^{n+1}}{3 n+6 (-1)^n}=frac{log(2)-1}{3}}$$
Are my calculations correct?
calculus sequences-and-series limits proof-verification closed-form
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
add a comment |
Let $displaystyle sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}$, does it converge or does it diverge?
I'm not completely sure that my calculation is correct, check it please.
$$begin{align}sum_{nge 0} frac{(-1)^{n+1}}{3 n+6 (-1)^n}&=-frac13sum_{nge 0} frac{(-1)^{n}}{n+2 (-1)^n}cdotfrac{(-1)^n}{(-1)^n}\&=-frac13sum_{nge 0} frac1{n(-1)^n+2}\&=-frac13lim_{mtoinfty}left(sum_{n=0\ 2mid n}^mfrac1{n+2}+sum_{n=0\ 2nmid n}^mfrac1{-n+2}right)\&=-frac13lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)end{align}$$
If $m$ is even then
$$begin{align}lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)&=lim_{mtoinfty}left(left(sum_{n=1}^{m-2}frac{(-1)^{n}}nright)+frac1m+frac1{m+2}+1right)\&=-log (2)+1end{align}$$
If $m$ is odd then
$$lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)=lim_{mtoinfty}left(left(sum_{n=1}^{m-1}frac{(-1)^{n}}nright)+frac1{m+1}+1right)=-log (2)+1$$
Then finally
$$bbox[2pt,border:yellow solid 2px]{sum_{n=0}^inftyfrac{(-1)^{n+1}}{3 n+6 (-1)^n}=frac{log(2)-1}{3}}$$
Are my calculations correct?
calculus sequences-and-series limits proof-verification closed-form
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
add a comment |
Let $displaystyle sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}$, does it converge or does it diverge?
I'm not completely sure that my calculation is correct, check it please.
$$begin{align}sum_{nge 0} frac{(-1)^{n+1}}{3 n+6 (-1)^n}&=-frac13sum_{nge 0} frac{(-1)^{n}}{n+2 (-1)^n}cdotfrac{(-1)^n}{(-1)^n}\&=-frac13sum_{nge 0} frac1{n(-1)^n+2}\&=-frac13lim_{mtoinfty}left(sum_{n=0\ 2mid n}^mfrac1{n+2}+sum_{n=0\ 2nmid n}^mfrac1{-n+2}right)\&=-frac13lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)end{align}$$
If $m$ is even then
$$begin{align}lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)&=lim_{mtoinfty}left(left(sum_{n=1}^{m-2}frac{(-1)^{n}}nright)+frac1m+frac1{m+2}+1right)\&=-log (2)+1end{align}$$
If $m$ is odd then
$$lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)=lim_{mtoinfty}left(left(sum_{n=1}^{m-1}frac{(-1)^{n}}nright)+frac1{m+1}+1right)=-log (2)+1$$
Then finally
$$bbox[2pt,border:yellow solid 2px]{sum_{n=0}^inftyfrac{(-1)^{n+1}}{3 n+6 (-1)^n}=frac{log(2)-1}{3}}$$
Are my calculations correct?
calculus sequences-and-series limits proof-verification closed-form
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
Let $displaystyle sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}$, does it converge or does it diverge?
I'm not completely sure that my calculation is correct, check it please.
$$begin{align}sum_{nge 0} frac{(-1)^{n+1}}{3 n+6 (-1)^n}&=-frac13sum_{nge 0} frac{(-1)^{n}}{n+2 (-1)^n}cdotfrac{(-1)^n}{(-1)^n}\&=-frac13sum_{nge 0} frac1{n(-1)^n+2}\&=-frac13lim_{mtoinfty}left(sum_{n=0\ 2mid n}^mfrac1{n+2}+sum_{n=0\ 2nmid n}^mfrac1{-n+2}right)\&=-frac13lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)end{align}$$
If $m$ is even then
$$begin{align}lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)&=lim_{mtoinfty}left(left(sum_{n=1}^{m-2}frac{(-1)^{n}}nright)+frac1m+frac1{m+2}+1right)\&=-log (2)+1end{align}$$
If $m$ is odd then
$$lim_{mtoinfty}left(sum_{n=2\ 2mid n}^{m+2}frac1{n}-sum_{n=-2\ 2nmid n}^{m-2}frac1{n}right)=lim_{mtoinfty}left(left(sum_{n=1}^{m-1}frac{(-1)^{n}}nright)+frac1{m+1}+1right)=-log (2)+1$$
Then finally
$$bbox[2pt,border:yellow solid 2px]{sum_{n=0}^inftyfrac{(-1)^{n+1}}{3 n+6 (-1)^n}=frac{log(2)-1}{3}}$$
Are my calculations correct?
calculus sequences-and-series limits proof-verification closed-form
calculus sequences-and-series limits proof-verification closed-form
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
edited Nov 30 at 16:17
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
asked Oct 30 '16 at 8:58
Masacroso
12.8k41746
12.8k41746
add a comment |
add a comment |
2 Answers
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One may set
$$
S_N=sum_{n=0}^N frac{(-1)^{n+1}}{3 n+6 (-1)^n},quad Nge0, tag1
$$ giving
$$
begin{align}
S_{2N}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=1}^{N} frac{1}{6p-9}
\\S_{2N}&=-frac16+frac13log 2+psiBig(N-frac12 Big)-psileft(N+2 right) tag2
end{align}
$$ and, as $N to infty$, giving
$$
lim_{N to infty}S_{2N}=-frac13+frac13log 2, tag3
$$
where $psi$ denotes the standard digamma function.
Similarly,$$
begin{align}
S_{2N+1}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=0}^{N} frac{1}{6p-3}
\\&S_{2N+1}=-frac13+frac13log 2+psiBig(N+frac12 Big)-psileft(N+2 right)
end{align}
$$ and, as $N to infty$, it gives
$$
lim_{N to infty}S_{2N+1}=-frac13+frac13log 2.
$$
We thus have
$$
sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}=-frac13+frac13log 2
$$
in agreement with your calculations.
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
add a comment |
Lemma: Suppose $a_nto 0$ and $sum_{n=1}^{infty}(a_{2n-1}+a_{2n})$ converges. Then $sum_{n=1}^{infty} a_n$ converges.
Proof: Let $S_n$ denote the $n$th partial sum of $sum a_n.$ The hypotheses imply that $S_{2n}$ converges. Since $S_{2n+1} = S_{2n} + a_{2n+1}$ and $a_{2n+1}to 0,$ $S_{2n+1}$ also converges. This implies $S_n$ converges.
In our problem, we have $a_nto 0.$ We also have
$$a_{2n-1}+a_{2n} =frac{1}{3(2n-1) -6}-frac{1}{3(2n)+1) -6} = frac{15}{(6n-9)(6n+6)}.$$
Thus for large $n,$ $a_{2n-1}+a_{2n}$ is positive and on the order of $1/n^2.$ It follows that $sum (a_{2n-1}+a_{2n})$ converges. By the lemma, $sum a_n$ converges.
answered Nov 30 at 16:51
zhw.
71.5k43075
add a comment |
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2 Answers
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2 Answers
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One may set
$$
S_N=sum_{n=0}^N frac{(-1)^{n+1}}{3 n+6 (-1)^n},quad Nge0, tag1
$$ giving
$$
begin{align}
S_{2N}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=1}^{N} frac{1}{6p-9}
\\S_{2N}&=-frac16+frac13log 2+psiBig(N-frac12 Big)-psileft(N+2 right) tag2
end{align}
$$ and, as $N to infty$, giving
$$
lim_{N to infty}S_{2N}=-frac13+frac13log 2, tag3
$$
where $psi$ denotes the standard digamma function.
Similarly,$$
begin{align}
S_{2N+1}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=0}^{N} frac{1}{6p-3}
\\&S_{2N+1}=-frac13+frac13log 2+psiBig(N+frac12 Big)-psileft(N+2 right)
end{align}
$$ and, as $N to infty$, it gives
$$
lim_{N to infty}S_{2N+1}=-frac13+frac13log 2.
$$
We thus have
$$
sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}=-frac13+frac13log 2
$$
in agreement with your calculations.
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
add a comment |
One may set
$$
S_N=sum_{n=0}^N frac{(-1)^{n+1}}{3 n+6 (-1)^n},quad Nge0, tag1
$$ giving
$$
begin{align}
S_{2N}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=1}^{N} frac{1}{6p-9}
\\S_{2N}&=-frac16+frac13log 2+psiBig(N-frac12 Big)-psileft(N+2 right) tag2
end{align}
$$ and, as $N to infty$, giving
$$
lim_{N to infty}S_{2N}=-frac13+frac13log 2, tag3
$$
where $psi$ denotes the standard digamma function.
Similarly,$$
begin{align}
S_{2N+1}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=0}^{N} frac{1}{6p-3}
\\&S_{2N+1}=-frac13+frac13log 2+psiBig(N+frac12 Big)-psileft(N+2 right)
end{align}
$$ and, as $N to infty$, it gives
$$
lim_{N to infty}S_{2N+1}=-frac13+frac13log 2.
$$
We thus have
$$
sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}=-frac13+frac13log 2
$$
in agreement with your calculations.
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
add a comment |
One may set
$$
S_N=sum_{n=0}^N frac{(-1)^{n+1}}{3 n+6 (-1)^n},quad Nge0, tag1
$$ giving
$$
begin{align}
S_{2N}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=1}^{N} frac{1}{6p-9}
\\S_{2N}&=-frac16+frac13log 2+psiBig(N-frac12 Big)-psileft(N+2 right) tag2
end{align}
$$ and, as $N to infty$, giving
$$
lim_{N to infty}S_{2N}=-frac13+frac13log 2, tag3
$$
where $psi$ denotes the standard digamma function.
Similarly,$$
begin{align}
S_{2N+1}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=0}^{N} frac{1}{6p-3}
\\&S_{2N+1}=-frac13+frac13log 2+psiBig(N+frac12 Big)-psileft(N+2 right)
end{align}
$$ and, as $N to infty$, it gives
$$
lim_{N to infty}S_{2N+1}=-frac13+frac13log 2.
$$
We thus have
$$
sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}=-frac13+frac13log 2
$$
in agreement with your calculations.
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
One may set
$$
S_N=sum_{n=0}^N frac{(-1)^{n+1}}{3 n+6 (-1)^n},quad Nge0, tag1
$$ giving
$$
begin{align}
S_{2N}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=1}^{N} frac{1}{6p-9}
\\S_{2N}&=-frac16+frac13log 2+psiBig(N-frac12 Big)-psileft(N+2 right) tag2
end{align}
$$ and, as $N to infty$, giving
$$
lim_{N to infty}S_{2N}=-frac13+frac13log 2, tag3
$$
where $psi$ denotes the standard digamma function.
Similarly,$$
begin{align}
S_{2N+1}&=sum_{p=0}^N frac{-1}{6p+6}+sum_{p=0}^{N} frac{1}{6p-3}
\\&S_{2N+1}=-frac13+frac13log 2+psiBig(N+frac12 Big)-psileft(N+2 right)
end{align}
$$ and, as $N to infty$, it gives
$$
lim_{N to infty}S_{2N+1}=-frac13+frac13log 2.
$$
We thus have
$$
sum_{n=0}^infty frac{(-1)^{n+1}}{3 n+6 (-1)^n}=-frac13+frac13log 2
$$
in agreement with your calculations.
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
edited Oct 30 '16 at 17:07
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
answered Oct 30 '16 at 10:55
Olivier Oloa
107k17175293
107k17175293
add a comment |
add a comment |
Lemma: Suppose $a_nto 0$ and $sum_{n=1}^{infty}(a_{2n-1}+a_{2n})$ converges. Then $sum_{n=1}^{infty} a_n$ converges.
Proof: Let $S_n$ denote the $n$th partial sum of $sum a_n.$ The hypotheses imply that $S_{2n}$ converges. Since $S_{2n+1} = S_{2n} + a_{2n+1}$ and $a_{2n+1}to 0,$ $S_{2n+1}$ also converges. This implies $S_n$ converges.
In our problem, we have $a_nto 0.$ We also have
$$a_{2n-1}+a_{2n} =frac{1}{3(2n-1) -6}-frac{1}{3(2n)+1) -6} = frac{15}{(6n-9)(6n+6)}.$$
Thus for large $n,$ $a_{2n-1}+a_{2n}$ is positive and on the order of $1/n^2.$ It follows that $sum (a_{2n-1}+a_{2n})$ converges. By the lemma, $sum a_n$ converges.
answered Nov 30 at 16:51
zhw.
71.5k43075
add a comment |
Lemma: Suppose $a_nto 0$ and $sum_{n=1}^{infty}(a_{2n-1}+a_{2n})$ converges. Then $sum_{n=1}^{infty} a_n$ converges.
Proof: Let $S_n$ denote the $n$th partial sum of $sum a_n.$ The hypotheses imply that $S_{2n}$ converges. Since $S_{2n+1} = S_{2n} + a_{2n+1}$ and $a_{2n+1}to 0,$ $S_{2n+1}$ also converges. This implies $S_n$ converges.
In our problem, we have $a_nto 0.$ We also have
$$a_{2n-1}+a_{2n} =frac{1}{3(2n-1) -6}-frac{1}{3(2n)+1) -6} = frac{15}{(6n-9)(6n+6)}.$$
Thus for large $n,$ $a_{2n-1}+a_{2n}$ is positive and on the order of $1/n^2.$ It follows that $sum (a_{2n-1}+a_{2n})$ converges. By the lemma, $sum a_n$ converges.
answered Nov 30 at 16:51
zhw.
71.5k43075
add a comment |
Lemma: Suppose $a_nto 0$ and $sum_{n=1}^{infty}(a_{2n-1}+a_{2n})$ converges. Then $sum_{n=1}^{infty} a_n$ converges.
Proof: Let $S_n$ denote the $n$th partial sum of $sum a_n.$ The hypotheses imply that $S_{2n}$ converges. Since $S_{2n+1} = S_{2n} + a_{2n+1}$ and $a_{2n+1}to 0,$ $S_{2n+1}$ also converges. This implies $S_n$ converges.
In our problem, we have $a_nto 0.$ We also have
$$a_{2n-1}+a_{2n} =frac{1}{3(2n-1) -6}-frac{1}{3(2n)+1) -6} = frac{15}{(6n-9)(6n+6)}.$$
Thus for large $n,$ $a_{2n-1}+a_{2n}$ is positive and on the order of $1/n^2.$ It follows that $sum (a_{2n-1}+a_{2n})$ converges. By the lemma, $sum a_n$ converges.
answered Nov 30 at 16:51
zhw.
71.5k43075
Lemma: Suppose $a_nto 0$ and $sum_{n=1}^{infty}(a_{2n-1}+a_{2n})$ converges. Then $sum_{n=1}^{infty} a_n$ converges.
Proof: Let $S_n$ denote the $n$th partial sum of $sum a_n.$ The hypotheses imply that $S_{2n}$ converges. Since $S_{2n+1} = S_{2n} + a_{2n+1}$ and $a_{2n+1}to 0,$ $S_{2n+1}$ also converges. This implies $S_n$ converges.
In our problem, we have $a_nto 0.$ We also have
$$a_{2n-1}+a_{2n} =frac{1}{3(2n-1) -6}-frac{1}{3(2n)+1) -6} = frac{15}{(6n-9)(6n+6)}.$$
Thus for large $n,$ $a_{2n-1}+a_{2n}$ is positive and on the order of $1/n^2.$ It follows that $sum (a_{2n-1}+a_{2n})$ converges. By the lemma, $sum a_n$ converges.
answered Nov 30 at 16:51
zhw.
71.5k43075
answered Nov 30 at 16:51
zhw.
71.5k43075
answered Nov 30 at 16:51
zhw.
71.5k43075
answered Nov 30 at 16:51
zhw.
71.5k43075
71.5k43075
add a comment |
add a comment |
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