Mixing
Investigate the convergence of $ sum_{n=1}^{infty } (-1)^{n}frac{n+2}{n(n+1)} $
$begingroup$
I am supposed to investigate the convergence of $ sum_{n=1}^{infty } (-1)^{n}frac{n+2}{n(n+1)} $. I'm unsure whether to use Leibniz' criterion or a comparison test and I really can't start. Thanks
convergence
edited Jan 25 at 16:59
Bernard
123k741116
asked Jan 25 at 16:55
J. DoeJ. Doe
61
$endgroup$
add a comment |
$begingroup$
I am supposed to investigate the convergence of $ sum_{n=1}^{infty } (-1)^{n}frac{n+2}{n(n+1)} $. I'm unsure whether to use Leibniz' criterion or a comparison test and I really can't start. Thanks
convergence
edited Jan 25 at 16:59
Bernard
123k741116
asked Jan 25 at 16:55
J. DoeJ. Doe
61
$endgroup$
$begingroup$
Alternating signs and decreasing magnitudes. So certainly convergent. Decreases only as $1/n$, so not absolutely convergent.
$endgroup$
– mjqxxxx
Jan 25 at 16:57
$begingroup$
What @mjqxxxx suggests works. The only "hard" part is to show the fact that the absolute value is indeed decreasing.
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– Clement C.
Jan 25 at 17:09
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After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– Clement C.
Jan 28 at 3:16
add a comment |
$begingroup$
I am supposed to investigate the convergence of $ sum_{n=1}^{infty } (-1)^{n}frac{n+2}{n(n+1)} $. I'm unsure whether to use Leibniz' criterion or a comparison test and I really can't start. Thanks
convergence
edited Jan 25 at 16:59
Bernard
123k741116
asked Jan 25 at 16:55
J. DoeJ. Doe
61
$endgroup$
I am supposed to investigate the convergence of $ sum_{n=1}^{infty } (-1)^{n}frac{n+2}{n(n+1)} $. I'm unsure whether to use Leibniz' criterion or a comparison test and I really can't start. Thanks
convergence
convergence
edited Jan 25 at 16:59
Bernard
123k741116
asked Jan 25 at 16:55
J. DoeJ. Doe
61
edited Jan 25 at 16:59
Bernard
123k741116
asked Jan 25 at 16:55
J. DoeJ. Doe
61
edited Jan 25 at 16:59
Bernard
123k741116
edited Jan 25 at 16:59
Bernard
123k741116
edited Jan 25 at 16:59
Bernard
123k741116
123k741116
asked Jan 25 at 16:55
J. DoeJ. Doe
61
asked Jan 25 at 16:55
J. DoeJ. Doe
61
asked Jan 25 at 16:55
J. DoeJ. Doe
61
61
$begingroup$
Alternating signs and decreasing magnitudes. So certainly convergent. Decreases only as $1/n$, so not absolutely convergent.
$endgroup$
– mjqxxxx
Jan 25 at 16:57
$begingroup$
What @mjqxxxx suggests works. The only "hard" part is to show the fact that the absolute value is indeed decreasing.
$endgroup$
– Clement C.
Jan 25 at 17:09
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– Clement C.
Jan 28 at 3:16
add a comment |
$begingroup$
Alternating signs and decreasing magnitudes. So certainly convergent. Decreases only as $1/n$, so not absolutely convergent.
$endgroup$
– mjqxxxx
Jan 25 at 16:57
$begingroup$
What @mjqxxxx suggests works. The only "hard" part is to show the fact that the absolute value is indeed decreasing.
$endgroup$
– Clement C.
Jan 25 at 17:09
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– Clement C.
Jan 28 at 3:16
$begingroup$
Alternating signs and decreasing magnitudes. So certainly convergent. Decreases only as $1/n$, so not absolutely convergent.
$endgroup$
– mjqxxxx
Jan 25 at 16:57
$begingroup$
Alternating signs and decreasing magnitudes. So certainly convergent. Decreases only as $1/n$, so not absolutely convergent.
$endgroup$
– mjqxxxx
Jan 25 at 16:57
$begingroup$
What @mjqxxxx suggests works. The only "hard" part is to show the fact that the absolute value is indeed decreasing.
$endgroup$
– Clement C.
Jan 25 at 17:09
$begingroup$
What @mjqxxxx suggests works. The only "hard" part is to show the fact that the absolute value is indeed decreasing.
$endgroup$
– Clement C.
Jan 25 at 17:09
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– Clement C.
Jan 28 at 3:16
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– Clement C.
Jan 28 at 3:16
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can directly use Leibniz' criterion, but you have to show the absolute value of the general term is non-increasing. Here, I suggest an (arguably) simpler way to see what's happening and prove convergence:
You have
$$
frac{n+2}{n(n+1)} = frac{1}{n+1} + frac{2}{n(n+1)}
$$
and therefore
$$
sum_{n=1}^infty (-1)^n frac{n+2}{n(n+1)} = sum_{n=1}^infty frac{(-1)^n}{n+1} + 2sum_{n=1}^infty frac{(-1)^n}{n(n+1)}
$$
The first series on the right-hand-side converges conditionally by Leibniz's criterion, the second converges absolutely.
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
$endgroup$
$begingroup$
Note: this type of approach works even when the general term is not actually amenable to Leibniz' criterion (or it's hard to show), but can be decomposed in the sum of a term that is and a term that is absolutely convergent. For instance (silly example), things of the sort $$sum_{n=1}^infty frac{(-1)^n}{n+(-1)^n sqrt{100n}}$$
$endgroup$
– Clement C.
Jan 25 at 17:05
add a comment |
$begingroup$
Just to show the longer, more formulaic approach:
Let $a_n: = frac{n+2}{n(n+1)}$. Then $$a_{n+1} - a_n = frac{n+3}{(n+1)(n+2)} - frac{n+2}{n(n+1)}$$ Using a common denominator we have $$a_{n+1}-a_n = frac{n^2 + 3n - n^2 - 4n - 4}{n(n+1)(n+2)} = frac{-(n+4)}{n(n+1)(n+2)}$$ and then clearly $$a_{n+1}-a_n = frac{-(n+4)}{n(n+1)(n+2)} < 0 hspace{2mm} forall n in mathbb{N}$$
So $a_n$ is a decreasing sequence. Furthermore, it is clear that $a_n rightarrow 0$ as $n rightarrow infty$, so by the alternating series test (or Leibniz criterion), the sum $sum_{n=1}^{infty} (-1)^n a_n$ converges.
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
$endgroup$
add a comment |
$begingroup$
1) $a_n = dfrac{n+2}{n(n+1)}= dfrac{1}{n} +dfrac{1}{n(n+1)}$.
$a_{n+1}=dfrac{1}{n+1}+ dfrac{1}{(n+1)(n+2)};$
Hence
$a_n > a_{n+1}$ (Why?).
2) $lim_{ n rightarrow infty}a_n=$
$lim_{n rightarrow infty}(dfrac{1}{n}+dfrac{1}{n(n+1)})=0$(why?).
Apply Leibniz criterion.
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
add a comment |
$begingroup$
Hint:
$$
{{n + 2} over {nleft( {n + 1} right)}} - {{n + 3} over {left( {n + 1} right)left( {n + 2} right)}} = {{n + 4} over {nleft( {n + 1} right)left( {n + 2} right)}}
$$
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can directly use Leibniz' criterion, but you have to show the absolute value of the general term is non-increasing. Here, I suggest an (arguably) simpler way to see what's happening and prove convergence:
You have
$$
frac{n+2}{n(n+1)} = frac{1}{n+1} + frac{2}{n(n+1)}
$$
and therefore
$$
sum_{n=1}^infty (-1)^n frac{n+2}{n(n+1)} = sum_{n=1}^infty frac{(-1)^n}{n+1} + 2sum_{n=1}^infty frac{(-1)^n}{n(n+1)}
$$
The first series on the right-hand-side converges conditionally by Leibniz's criterion, the second converges absolutely.
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
$endgroup$
$begingroup$
Note: this type of approach works even when the general term is not actually amenable to Leibniz' criterion (or it's hard to show), but can be decomposed in the sum of a term that is and a term that is absolutely convergent. For instance (silly example), things of the sort $$sum_{n=1}^infty frac{(-1)^n}{n+(-1)^n sqrt{100n}}$$
$endgroup$
– Clement C.
Jan 25 at 17:05
add a comment |
$begingroup$
You can directly use Leibniz' criterion, but you have to show the absolute value of the general term is non-increasing. Here, I suggest an (arguably) simpler way to see what's happening and prove convergence:
You have
$$
frac{n+2}{n(n+1)} = frac{1}{n+1} + frac{2}{n(n+1)}
$$
and therefore
$$
sum_{n=1}^infty (-1)^n frac{n+2}{n(n+1)} = sum_{n=1}^infty frac{(-1)^n}{n+1} + 2sum_{n=1}^infty frac{(-1)^n}{n(n+1)}
$$
The first series on the right-hand-side converges conditionally by Leibniz's criterion, the second converges absolutely.
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
$endgroup$
$begingroup$
Note: this type of approach works even when the general term is not actually amenable to Leibniz' criterion (or it's hard to show), but can be decomposed in the sum of a term that is and a term that is absolutely convergent. For instance (silly example), things of the sort $$sum_{n=1}^infty frac{(-1)^n}{n+(-1)^n sqrt{100n}}$$
$endgroup$
– Clement C.
Jan 25 at 17:05
add a comment |
$begingroup$
You can directly use Leibniz' criterion, but you have to show the absolute value of the general term is non-increasing. Here, I suggest an (arguably) simpler way to see what's happening and prove convergence:
You have
$$
frac{n+2}{n(n+1)} = frac{1}{n+1} + frac{2}{n(n+1)}
$$
and therefore
$$
sum_{n=1}^infty (-1)^n frac{n+2}{n(n+1)} = sum_{n=1}^infty frac{(-1)^n}{n+1} + 2sum_{n=1}^infty frac{(-1)^n}{n(n+1)}
$$
The first series on the right-hand-side converges conditionally by Leibniz's criterion, the second converges absolutely.
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
$endgroup$
You can directly use Leibniz' criterion, but you have to show the absolute value of the general term is non-increasing. Here, I suggest an (arguably) simpler way to see what's happening and prove convergence:
You have
$$
frac{n+2}{n(n+1)} = frac{1}{n+1} + frac{2}{n(n+1)}
$$
and therefore
$$
sum_{n=1}^infty (-1)^n frac{n+2}{n(n+1)} = sum_{n=1}^infty frac{(-1)^n}{n+1} + 2sum_{n=1}^infty frac{(-1)^n}{n(n+1)}
$$
The first series on the right-hand-side converges conditionally by Leibniz's criterion, the second converges absolutely.
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
edited Jan 25 at 17:36
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
answered Jan 25 at 16:59
Clement C.Clement C.
50.9k33992
50.9k33992
$begingroup$
Note: this type of approach works even when the general term is not actually amenable to Leibniz' criterion (or it's hard to show), but can be decomposed in the sum of a term that is and a term that is absolutely convergent. For instance (silly example), things of the sort $$sum_{n=1}^infty frac{(-1)^n}{n+(-1)^n sqrt{100n}}$$
$endgroup$
– Clement C.
Jan 25 at 17:05
add a comment |
$begingroup$
Note: this type of approach works even when the general term is not actually amenable to Leibniz' criterion (or it's hard to show), but can be decomposed in the sum of a term that is and a term that is absolutely convergent. For instance (silly example), things of the sort $$sum_{n=1}^infty frac{(-1)^n}{n+(-1)^n sqrt{100n}}$$
$endgroup$
– Clement C.
Jan 25 at 17:05
$begingroup$
Note: this type of approach works even when the general term is not actually amenable to Leibniz' criterion (or it's hard to show), but can be decomposed in the sum of a term that is and a term that is absolutely convergent. For instance (silly example), things of the sort $$sum_{n=1}^infty frac{(-1)^n}{n+(-1)^n sqrt{100n}}$$
$endgroup$
– Clement C.
Jan 25 at 17:05
$begingroup$
Note: this type of approach works even when the general term is not actually amenable to Leibniz' criterion (or it's hard to show), but can be decomposed in the sum of a term that is and a term that is absolutely convergent. For instance (silly example), things of the sort $$sum_{n=1}^infty frac{(-1)^n}{n+(-1)^n sqrt{100n}}$$
$endgroup$
– Clement C.
Jan 25 at 17:05
add a comment |
$begingroup$
Just to show the longer, more formulaic approach:
Let $a_n: = frac{n+2}{n(n+1)}$. Then $$a_{n+1} - a_n = frac{n+3}{(n+1)(n+2)} - frac{n+2}{n(n+1)}$$ Using a common denominator we have $$a_{n+1}-a_n = frac{n^2 + 3n - n^2 - 4n - 4}{n(n+1)(n+2)} = frac{-(n+4)}{n(n+1)(n+2)}$$ and then clearly $$a_{n+1}-a_n = frac{-(n+4)}{n(n+1)(n+2)} < 0 hspace{2mm} forall n in mathbb{N}$$
So $a_n$ is a decreasing sequence. Furthermore, it is clear that $a_n rightarrow 0$ as $n rightarrow infty$, so by the alternating series test (or Leibniz criterion), the sum $sum_{n=1}^{infty} (-1)^n a_n$ converges.
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
$endgroup$
add a comment |
$begingroup$
Just to show the longer, more formulaic approach:
Let $a_n: = frac{n+2}{n(n+1)}$. Then $$a_{n+1} - a_n = frac{n+3}{(n+1)(n+2)} - frac{n+2}{n(n+1)}$$ Using a common denominator we have $$a_{n+1}-a_n = frac{n^2 + 3n - n^2 - 4n - 4}{n(n+1)(n+2)} = frac{-(n+4)}{n(n+1)(n+2)}$$ and then clearly $$a_{n+1}-a_n = frac{-(n+4)}{n(n+1)(n+2)} < 0 hspace{2mm} forall n in mathbb{N}$$
So $a_n$ is a decreasing sequence. Furthermore, it is clear that $a_n rightarrow 0$ as $n rightarrow infty$, so by the alternating series test (or Leibniz criterion), the sum $sum_{n=1}^{infty} (-1)^n a_n$ converges.
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
$endgroup$
add a comment |
$begingroup$
Just to show the longer, more formulaic approach:
Let $a_n: = frac{n+2}{n(n+1)}$. Then $$a_{n+1} - a_n = frac{n+3}{(n+1)(n+2)} - frac{n+2}{n(n+1)}$$ Using a common denominator we have $$a_{n+1}-a_n = frac{n^2 + 3n - n^2 - 4n - 4}{n(n+1)(n+2)} = frac{-(n+4)}{n(n+1)(n+2)}$$ and then clearly $$a_{n+1}-a_n = frac{-(n+4)}{n(n+1)(n+2)} < 0 hspace{2mm} forall n in mathbb{N}$$
So $a_n$ is a decreasing sequence. Furthermore, it is clear that $a_n rightarrow 0$ as $n rightarrow infty$, so by the alternating series test (or Leibniz criterion), the sum $sum_{n=1}^{infty} (-1)^n a_n$ converges.
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
$endgroup$
Just to show the longer, more formulaic approach:
Let $a_n: = frac{n+2}{n(n+1)}$. Then $$a_{n+1} - a_n = frac{n+3}{(n+1)(n+2)} - frac{n+2}{n(n+1)}$$ Using a common denominator we have $$a_{n+1}-a_n = frac{n^2 + 3n - n^2 - 4n - 4}{n(n+1)(n+2)} = frac{-(n+4)}{n(n+1)(n+2)}$$ and then clearly $$a_{n+1}-a_n = frac{-(n+4)}{n(n+1)(n+2)} < 0 hspace{2mm} forall n in mathbb{N}$$
So $a_n$ is a decreasing sequence. Furthermore, it is clear that $a_n rightarrow 0$ as $n rightarrow infty$, so by the alternating series test (or Leibniz criterion), the sum $sum_{n=1}^{infty} (-1)^n a_n$ converges.
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
answered Jan 25 at 17:21
PhysicsMathsLovePhysicsMathsLove
1,220414
1,220414
add a comment |
add a comment |
$begingroup$
1) $a_n = dfrac{n+2}{n(n+1)}= dfrac{1}{n} +dfrac{1}{n(n+1)}$.
$a_{n+1}=dfrac{1}{n+1}+ dfrac{1}{(n+1)(n+2)};$
Hence
$a_n > a_{n+1}$ (Why?).
2) $lim_{ n rightarrow infty}a_n=$
$lim_{n rightarrow infty}(dfrac{1}{n}+dfrac{1}{n(n+1)})=0$(why?).
Apply Leibniz criterion.
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
add a comment |
$begingroup$
1) $a_n = dfrac{n+2}{n(n+1)}= dfrac{1}{n} +dfrac{1}{n(n+1)}$.
$a_{n+1}=dfrac{1}{n+1}+ dfrac{1}{(n+1)(n+2)};$
Hence
$a_n > a_{n+1}$ (Why?).
2) $lim_{ n rightarrow infty}a_n=$
$lim_{n rightarrow infty}(dfrac{1}{n}+dfrac{1}{n(n+1)})=0$(why?).
Apply Leibniz criterion.
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
add a comment |
$begingroup$
1) $a_n = dfrac{n+2}{n(n+1)}= dfrac{1}{n} +dfrac{1}{n(n+1)}$.
$a_{n+1}=dfrac{1}{n+1}+ dfrac{1}{(n+1)(n+2)};$
Hence
$a_n > a_{n+1}$ (Why?).
2) $lim_{ n rightarrow infty}a_n=$
$lim_{n rightarrow infty}(dfrac{1}{n}+dfrac{1}{n(n+1)})=0$(why?).
Apply Leibniz criterion.
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
1) $a_n = dfrac{n+2}{n(n+1)}= dfrac{1}{n} +dfrac{1}{n(n+1)}$.
$a_{n+1}=dfrac{1}{n+1}+ dfrac{1}{(n+1)(n+2)};$
Hence
$a_n > a_{n+1}$ (Why?).
2) $lim_{ n rightarrow infty}a_n=$
$lim_{n rightarrow infty}(dfrac{1}{n}+dfrac{1}{n(n+1)})=0$(why?).
Apply Leibniz criterion.
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
answered Jan 25 at 17:21
Peter SzilasPeter Szilas
11.6k2822
11.6k2822
add a comment |
add a comment |
$begingroup$
Hint:
$$
{{n + 2} over {nleft( {n + 1} right)}} - {{n + 3} over {left( {n + 1} right)left( {n + 2} right)}} = {{n + 4} over {nleft( {n + 1} right)left( {n + 2} right)}}
$$
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Hint:
$$
{{n + 2} over {nleft( {n + 1} right)}} - {{n + 3} over {left( {n + 1} right)left( {n + 2} right)}} = {{n + 4} over {nleft( {n + 1} right)left( {n + 2} right)}}
$$
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Hint:
$$
{{n + 2} over {nleft( {n + 1} right)}} - {{n + 3} over {left( {n + 1} right)left( {n + 2} right)}} = {{n + 4} over {nleft( {n + 1} right)left( {n + 2} right)}}
$$
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
$endgroup$
Hint:
$$
{{n + 2} over {nleft( {n + 1} right)}} - {{n + 3} over {left( {n + 1} right)left( {n + 2} right)}} = {{n + 4} over {nleft( {n + 1} right)left( {n + 2} right)}}
$$
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
answered Jan 25 at 17:30
G CabG Cab
20.4k31341
20.4k31341
add a comment |
add a comment |
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$begingroup$
Alternating signs and decreasing magnitudes. So certainly convergent. Decreases only as $1/n$, so not absolutely convergent.
$endgroup$
– mjqxxxx
Jan 25 at 16:57
$begingroup$
What @mjqxxxx suggests works. The only "hard" part is to show the fact that the absolute value is indeed decreasing.
$endgroup$
– Clement C.
Jan 25 at 17:09
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– Clement C.
Jan 28 at 3:16