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compute $frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$
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compute the summation
$sum_ n=1^infty fracn3.5........(2n+1)= ?$
My attempts : i take $a_n =fracn3.5........(2n+1)$
Now =$frac12$ .$frac(2n+ 1) - 13.5........(2n+1)= frac12(frac13.5........(2n-1) -frac13.5........(2n+1))$
after that i can not able to proceed further,,,
pliz help me..
thanks u
real-analysis
asked Aug 4 at 0:27
stupid
4888
add a comment |Â
up vote
0
down vote
favorite
compute the summation
$sum_ n=1^infty fracn3.5........(2n+1)= ?$
My attempts : i take $a_n =fracn3.5........(2n+1)$
Now =$frac12$ .$frac(2n+ 1) - 13.5........(2n+1)= frac12(frac13.5........(2n-1) -frac13.5........(2n+1))$
after that i can not able to proceed further,,,
pliz help me..
thanks u
real-analysis
asked Aug 4 at 0:27
stupid
4888
2
Write a partial sum, delete the parentheses, and look at consecutive terms.
– spiralstotheleft
Aug 4 at 0:30
im not getting @spiralstotheleft
– stupid
Aug 4 at 0:35
A partial sum is $sum_n=1^Mfrac12left(frac13cdot5...(2n-1)-frac13cdot5...(2n+1)right)=frac12-frac13cdot5...(2M+1)$, which tends to $frac12$ since the last term tends to $0$. Therefore, the sum of your series is $frac12$. Note: Opening parenthesis before knowing that the partial sums converge, like it was done in your selected answer, is not justified. If you did it for another series like $sum_n=1^infty(1-1)$ you get a series with a different character.
– spiralstotheleft
2 days ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
compute the summation
$sum_ n=1^infty fracn3.5........(2n+1)= ?$
My attempts : i take $a_n =fracn3.5........(2n+1)$
Now =$frac12$ .$frac(2n+ 1) - 13.5........(2n+1)= frac12(frac13.5........(2n-1) -frac13.5........(2n+1))$
after that i can not able to proceed further,,,
pliz help me..
thanks u
real-analysis
asked Aug 4 at 0:27
stupid
4888
compute the summation
$sum_ n=1^infty fracn3.5........(2n+1)= ?$
My attempts : i take $a_n =fracn3.5........(2n+1)$
Now =$frac12$ .$frac(2n+ 1) - 13.5........(2n+1)= frac12(frac13.5........(2n-1) -frac13.5........(2n+1))$
after that i can not able to proceed further,,,
pliz help me..
thanks u
real-analysis
asked Aug 4 at 0:27
stupid
4888
asked Aug 4 at 0:27
stupid
4888
asked Aug 4 at 0:27
stupid
4888
asked Aug 4 at 0:27
stupid
4888
4888
2
Write a partial sum, delete the parentheses, and look at consecutive terms.
– spiralstotheleft
Aug 4 at 0:30
im not getting @spiralstotheleft
– stupid
Aug 4 at 0:35
A partial sum is $sum_n=1^Mfrac12left(frac13cdot5...(2n-1)-frac13cdot5...(2n+1)right)=frac12-frac13cdot5...(2M+1)$, which tends to $frac12$ since the last term tends to $0$. Therefore, the sum of your series is $frac12$. Note: Opening parenthesis before knowing that the partial sums converge, like it was done in your selected answer, is not justified. If you did it for another series like $sum_n=1^infty(1-1)$ you get a series with a different character.
– spiralstotheleft
2 days ago
add a comment |Â
2
Write a partial sum, delete the parentheses, and look at consecutive terms.
– spiralstotheleft
Aug 4 at 0:30
im not getting @spiralstotheleft
– stupid
Aug 4 at 0:35
A partial sum is $sum_n=1^Mfrac12left(frac13cdot5...(2n-1)-frac13cdot5...(2n+1)right)=frac12-frac13cdot5...(2M+1)$, which tends to $frac12$ since the last term tends to $0$. Therefore, the sum of your series is $frac12$. Note: Opening parenthesis before knowing that the partial sums converge, like it was done in your selected answer, is not justified. If you did it for another series like $sum_n=1^infty(1-1)$ you get a series with a different character.
– spiralstotheleft
2 days ago
2
2
Write a partial sum, delete the parentheses, and look at consecutive terms.
– spiralstotheleft
Aug 4 at 0:30
Write a partial sum, delete the parentheses, and look at consecutive terms.
– spiralstotheleft
Aug 4 at 0:30
im not getting @spiralstotheleft
– stupid
Aug 4 at 0:35
im not getting @spiralstotheleft
– stupid
Aug 4 at 0:35
A partial sum is $sum_n=1^Mfrac12left(frac13cdot5...(2n-1)-frac13cdot5...(2n+1)right)=frac12-frac13cdot5...(2M+1)$, which tends to $frac12$ since the last term tends to $0$. Therefore, the sum of your series is $frac12$. Note: Opening parenthesis before knowing that the partial sums converge, like it was done in your selected answer, is not justified. If you did it for another series like $sum_n=1^infty(1-1)$ you get a series with a different character.
– spiralstotheleft
2 days ago
A partial sum is $sum_n=1^Mfrac12left(frac13cdot5...(2n-1)-frac13cdot5...(2n+1)right)=frac12-frac13cdot5...(2M+1)$, which tends to $frac12$ since the last term tends to $0$. Therefore, the sum of your series is $frac12$. Note: Opening parenthesis before knowing that the partial sums converge, like it was done in your selected answer, is not justified. If you did it for another series like $sum_n=1^infty(1-1)$ you get a series with a different character.
– spiralstotheleft
2 days ago
add a comment |Â
2 Answers
2
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up vote
1
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accepted
Your work so far is good. Just keep going.
$$sum_n=1^infty a_n = sum_n=1^infty frac12 left(frac11 cdot 3 cdot 5 cdots (2n-1) - frac11 cdot 3 cdot 5 cdots (2n+1)right)
= frac12 left(frac11 - frac11 cdot 3 + frac11cdot 3 - frac11 cdot 3 cdot 5 + frac11 cdot 3 cdot 5 - frac11 cdot 3 cdot 5 cdot 7 + cdotsright)$$
answered Aug 4 at 1:09
angryavian
34.3k12874
thanks u @ angryavian can u write ur answer as detail as in previous question ? the value of ranks A., ImA and Ker A
– stupid
Aug 4 at 1:29
add a comment |Â
up vote
1
down vote
You don't need to compute $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$
You need to compute $sum_n=1^K a_n$.
Which is $frac 12(frac 11 - frac 13) + frac 12(frac 13-frac 13cdot 5)+.....+ frac12(frac13.5…(2K-3) -frac13.5…(2K-1)) + frac12(frac13.5…(2K-1) -frac13.5…(2K+1))$
And the compute $sum_n=1^infty a_n =limlimits_Kto inftysum_n=1^K a_n$
====
Note: If $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$ we can let $b_n =frac13.5…(2n-1)$
and then we have
$$sum_n=1^inftyfrac 12 (b_n - b_n+1)$$.
And you should be able so solve that even if you don't have any frigging idea what $b_n$ was defined as.
answered Aug 4 at 1:09
fleablood
60.1k22575
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your work so far is good. Just keep going.
$$sum_n=1^infty a_n = sum_n=1^infty frac12 left(frac11 cdot 3 cdot 5 cdots (2n-1) - frac11 cdot 3 cdot 5 cdots (2n+1)right)
= frac12 left(frac11 - frac11 cdot 3 + frac11cdot 3 - frac11 cdot 3 cdot 5 + frac11 cdot 3 cdot 5 - frac11 cdot 3 cdot 5 cdot 7 + cdotsright)$$
answered Aug 4 at 1:09
angryavian
34.3k12874
thanks u @ angryavian can u write ur answer as detail as in previous question ? the value of ranks A., ImA and Ker A
– stupid
Aug 4 at 1:29
add a comment |Â
up vote
1
down vote
accepted
Your work so far is good. Just keep going.
$$sum_n=1^infty a_n = sum_n=1^infty frac12 left(frac11 cdot 3 cdot 5 cdots (2n-1) - frac11 cdot 3 cdot 5 cdots (2n+1)right)
= frac12 left(frac11 - frac11 cdot 3 + frac11cdot 3 - frac11 cdot 3 cdot 5 + frac11 cdot 3 cdot 5 - frac11 cdot 3 cdot 5 cdot 7 + cdotsright)$$
answered Aug 4 at 1:09
angryavian
34.3k12874
thanks u @ angryavian can u write ur answer as detail as in previous question ? the value of ranks A., ImA and Ker A
– stupid
Aug 4 at 1:29
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your work so far is good. Just keep going.
$$sum_n=1^infty a_n = sum_n=1^infty frac12 left(frac11 cdot 3 cdot 5 cdots (2n-1) - frac11 cdot 3 cdot 5 cdots (2n+1)right)
= frac12 left(frac11 - frac11 cdot 3 + frac11cdot 3 - frac11 cdot 3 cdot 5 + frac11 cdot 3 cdot 5 - frac11 cdot 3 cdot 5 cdot 7 + cdotsright)$$
answered Aug 4 at 1:09
angryavian
34.3k12874
Your work so far is good. Just keep going.
$$sum_n=1^infty a_n = sum_n=1^infty frac12 left(frac11 cdot 3 cdot 5 cdots (2n-1) - frac11 cdot 3 cdot 5 cdots (2n+1)right)
= frac12 left(frac11 - frac11 cdot 3 + frac11cdot 3 - frac11 cdot 3 cdot 5 + frac11 cdot 3 cdot 5 - frac11 cdot 3 cdot 5 cdot 7 + cdotsright)$$
answered Aug 4 at 1:09
angryavian
34.3k12874
answered Aug 4 at 1:09
angryavian
34.3k12874
answered Aug 4 at 1:09
angryavian
34.3k12874
answered Aug 4 at 1:09
angryavian
34.3k12874
34.3k12874
thanks u @ angryavian can u write ur answer as detail as in previous question ? the value of ranks A., ImA and Ker A
– stupid
Aug 4 at 1:29
add a comment |Â
thanks u @ angryavian can u write ur answer as detail as in previous question ? the value of ranks A., ImA and Ker A
– stupid
Aug 4 at 1:29
thanks u @ angryavian can u write ur answer as detail as in previous question ? the value of ranks A., ImA and Ker A
– stupid
Aug 4 at 1:29
thanks u @ angryavian can u write ur answer as detail as in previous question ? the value of ranks A., ImA and Ker A
– stupid
Aug 4 at 1:29
add a comment |Â
up vote
1
down vote
You don't need to compute $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$
You need to compute $sum_n=1^K a_n$.
Which is $frac 12(frac 11 - frac 13) + frac 12(frac 13-frac 13cdot 5)+.....+ frac12(frac13.5…(2K-3) -frac13.5…(2K-1)) + frac12(frac13.5…(2K-1) -frac13.5…(2K+1))$
And the compute $sum_n=1^infty a_n =limlimits_Kto inftysum_n=1^K a_n$
====
Note: If $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$ we can let $b_n =frac13.5…(2n-1)$
and then we have
$$sum_n=1^inftyfrac 12 (b_n - b_n+1)$$.
And you should be able so solve that even if you don't have any frigging idea what $b_n$ was defined as.
answered Aug 4 at 1:09
fleablood
60.1k22575
add a comment |Â
up vote
1
down vote
You don't need to compute $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$
You need to compute $sum_n=1^K a_n$.
Which is $frac 12(frac 11 - frac 13) + frac 12(frac 13-frac 13cdot 5)+.....+ frac12(frac13.5…(2K-3) -frac13.5…(2K-1)) + frac12(frac13.5…(2K-1) -frac13.5…(2K+1))$
And the compute $sum_n=1^infty a_n =limlimits_Kto inftysum_n=1^K a_n$
====
Note: If $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$ we can let $b_n =frac13.5…(2n-1)$
and then we have
$$sum_n=1^inftyfrac 12 (b_n - b_n+1)$$.
And you should be able so solve that even if you don't have any frigging idea what $b_n$ was defined as.
answered Aug 4 at 1:09
fleablood
60.1k22575
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You don't need to compute $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$
You need to compute $sum_n=1^K a_n$.
Which is $frac 12(frac 11 - frac 13) + frac 12(frac 13-frac 13cdot 5)+.....+ frac12(frac13.5…(2K-3) -frac13.5…(2K-1)) + frac12(frac13.5…(2K-1) -frac13.5…(2K+1))$
And the compute $sum_n=1^infty a_n =limlimits_Kto inftysum_n=1^K a_n$
====
Note: If $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$ we can let $b_n =frac13.5…(2n-1)$
and then we have
$$sum_n=1^inftyfrac 12 (b_n - b_n+1)$$.
And you should be able so solve that even if you don't have any frigging idea what $b_n$ was defined as.
answered Aug 4 at 1:09
fleablood
60.1k22575
You don't need to compute $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$
You need to compute $sum_n=1^K a_n$.
Which is $frac 12(frac 11 - frac 13) + frac 12(frac 13-frac 13cdot 5)+.....+ frac12(frac13.5…(2K-3) -frac13.5…(2K-1)) + frac12(frac13.5…(2K-1) -frac13.5…(2K+1))$
And the compute $sum_n=1^infty a_n =limlimits_Kto inftysum_n=1^K a_n$
====
Note: If $a_n = frac12(frac13.5…(2n-1) -frac13.5…(2n+1))$ we can let $b_n =frac13.5…(2n-1)$
and then we have
$$sum_n=1^inftyfrac 12 (b_n - b_n+1)$$.
And you should be able so solve that even if you don't have any frigging idea what $b_n$ was defined as.
answered Aug 4 at 1:09
fleablood
60.1k22575
edited Aug 4 at 1:14
answered Aug 4 at 1:09
fleablood
60.1k22575
answered Aug 4 at 1:09
fleablood
60.1k22575
answered Aug 4 at 1:09
fleablood
60.1k22575
60.1k22575
add a comment |Â
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2
Write a partial sum, delete the parentheses, and look at consecutive terms.
– spiralstotheleft
Aug 4 at 0:30
im not getting @spiralstotheleft
– stupid
Aug 4 at 0:35
A partial sum is $sum_n=1^Mfrac12left(frac13cdot5...(2n-1)-frac13cdot5...(2n+1)right)=frac12-frac13cdot5...(2M+1)$, which tends to $frac12$ since the last term tends to $0$. Therefore, the sum of your series is $frac12$. Note: Opening parenthesis before knowing that the partial sums converge, like it was done in your selected answer, is not justified. If you did it for another series like $sum_n=1^infty(1-1)$ you get a series with a different character.
– spiralstotheleft
2 days ago