Anyone knows how to calculate the sum of this series?
up vote
1
down vote
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$$ sum_{n=1}^{999} log_{10}left(frac{n+1}{n}right) $$
Can anybody help me how to calculate this summation?
sequences-and-series summation logarithms taylor-expansion euler-maclaurin
edited Nov 18 at 18:04
NickD
9731412
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
|
show 2 more comments
up vote
1
down vote
favorite
$$ sum_{n=1}^{999} log_{10}left(frac{n+1}{n}right) $$
Can anybody help me how to calculate this summation?
sequences-and-series summation logarithms taylor-expansion euler-maclaurin
edited Nov 18 at 18:04
NickD
9731412
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
Sorry this is the first time I'm using this so I had no idea how to properly put an equation so there is an image
– Minh Nguyen Nhat
Nov 18 at 0:55
Hint: logarithms turn products into sums.
– NickD
Nov 18 at 1:01
Yeah I figured that part and got log10(1000!) - log10(999!) but don't know how to derive that one without using wolfram alpha
– Minh Nguyen Nhat
Nov 18 at 1:03
Keep it as a product: there's lots of cancellations.
– NickD
Nov 18 at 1:05
1
Ahh, I see what you mean! Thanks very much!
– Minh Nguyen Nhat
Nov 18 at 1:09
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$ sum_{n=1}^{999} log_{10}left(frac{n+1}{n}right) $$
Can anybody help me how to calculate this summation?
sequences-and-series summation logarithms taylor-expansion euler-maclaurin
edited Nov 18 at 18:04
NickD
9731412
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
$$ sum_{n=1}^{999} log_{10}left(frac{n+1}{n}right) $$
Can anybody help me how to calculate this summation?
sequences-and-series summation logarithms taylor-expansion euler-maclaurin
sequences-and-series summation logarithms taylor-expansion euler-maclaurin
edited Nov 18 at 18:04
NickD
9731412
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
edited Nov 18 at 18:04
NickD
9731412
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
edited Nov 18 at 18:04
NickD
9731412
edited Nov 18 at 18:04
NickD
9731412
edited Nov 18 at 18:04
NickD
9731412
9731412
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
asked Nov 18 at 0:54
Minh Nguyen Nhat
133
133
Sorry this is the first time I'm using this so I had no idea how to properly put an equation so there is an image
– Minh Nguyen Nhat
Nov 18 at 0:55
Hint: logarithms turn products into sums.
– NickD
Nov 18 at 1:01
Yeah I figured that part and got log10(1000!) - log10(999!) but don't know how to derive that one without using wolfram alpha
– Minh Nguyen Nhat
Nov 18 at 1:03
Keep it as a product: there's lots of cancellations.
– NickD
Nov 18 at 1:05
1
Ahh, I see what you mean! Thanks very much!
– Minh Nguyen Nhat
Nov 18 at 1:09
|
show 2 more comments
Sorry this is the first time I'm using this so I had no idea how to properly put an equation so there is an image
– Minh Nguyen Nhat
Nov 18 at 0:55
Hint: logarithms turn products into sums.
– NickD
Nov 18 at 1:01
Yeah I figured that part and got log10(1000!) - log10(999!) but don't know how to derive that one without using wolfram alpha
– Minh Nguyen Nhat
Nov 18 at 1:03
Keep it as a product: there's lots of cancellations.
– NickD
Nov 18 at 1:05
1
Ahh, I see what you mean! Thanks very much!
– Minh Nguyen Nhat
Nov 18 at 1:09
Sorry this is the first time I'm using this so I had no idea how to properly put an equation so there is an image
– Minh Nguyen Nhat
Nov 18 at 0:55
Sorry this is the first time I'm using this so I had no idea how to properly put an equation so there is an image
– Minh Nguyen Nhat
Nov 18 at 0:55
Hint: logarithms turn products into sums.
– NickD
Nov 18 at 1:01
Hint: logarithms turn products into sums.
– NickD
Nov 18 at 1:01
Yeah I figured that part and got log10(1000!) - log10(999!) but don't know how to derive that one without using wolfram alpha
– Minh Nguyen Nhat
Nov 18 at 1:03
Yeah I figured that part and got log10(1000!) - log10(999!) but don't know how to derive that one without using wolfram alpha
– Minh Nguyen Nhat
Nov 18 at 1:03
Keep it as a product: there's lots of cancellations.
– NickD
Nov 18 at 1:05
Keep it as a product: there's lots of cancellations.
– NickD
Nov 18 at 1:05
1
1
Ahh, I see what you mean! Thanks very much!
– Minh Nguyen Nhat
Nov 18 at 1:09
Ahh, I see what you mean! Thanks very much!
– Minh Nguyen Nhat
Nov 18 at 1:09
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You were on the right track! Like Minh Nguyen Nhat, you have to take $log(1000!)-log(999!)$ as $log(1000!/999!)$ to get the answer.
$log(2/1)+log(3/2)+log(4/3)+$...$+log(1000/999)=log(1000!/999!)=log(1000)=3$
edited Nov 18 at 1:42
Robson
751221
answered Nov 18 at 1:12
Muchang Bahng
562
1
Yeah I figured it out thank you!!!
– Minh Nguyen Nhat
Nov 18 at 1:15
1
hello @Muchang Bahng, you can use MathJax to improve your answers and help more people in a more beatiful way here is a link that could be useful for you math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 18 at 1:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You were on the right track! Like Minh Nguyen Nhat, you have to take $log(1000!)-log(999!)$ as $log(1000!/999!)$ to get the answer.
$log(2/1)+log(3/2)+log(4/3)+$...$+log(1000/999)=log(1000!/999!)=log(1000)=3$
edited Nov 18 at 1:42
Robson
751221
answered Nov 18 at 1:12
Muchang Bahng
562
1
Yeah I figured it out thank you!!!
– Minh Nguyen Nhat
Nov 18 at 1:15
1
hello @Muchang Bahng, you can use MathJax to improve your answers and help more people in a more beatiful way here is a link that could be useful for you math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 18 at 1:29
add a comment |
up vote
1
down vote
accepted
You were on the right track! Like Minh Nguyen Nhat, you have to take $log(1000!)-log(999!)$ as $log(1000!/999!)$ to get the answer.
$log(2/1)+log(3/2)+log(4/3)+$...$+log(1000/999)=log(1000!/999!)=log(1000)=3$
edited Nov 18 at 1:42
Robson
751221
answered Nov 18 at 1:12
Muchang Bahng
562
1
Yeah I figured it out thank you!!!
– Minh Nguyen Nhat
Nov 18 at 1:15
1
hello @Muchang Bahng, you can use MathJax to improve your answers and help more people in a more beatiful way here is a link that could be useful for you math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 18 at 1:29
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You were on the right track! Like Minh Nguyen Nhat, you have to take $log(1000!)-log(999!)$ as $log(1000!/999!)$ to get the answer.
$log(2/1)+log(3/2)+log(4/3)+$...$+log(1000/999)=log(1000!/999!)=log(1000)=3$
edited Nov 18 at 1:42
Robson
751221
answered Nov 18 at 1:12
Muchang Bahng
562
You were on the right track! Like Minh Nguyen Nhat, you have to take $log(1000!)-log(999!)$ as $log(1000!/999!)$ to get the answer.
$log(2/1)+log(3/2)+log(4/3)+$...$+log(1000/999)=log(1000!/999!)=log(1000)=3$
edited Nov 18 at 1:42
Robson
751221
answered Nov 18 at 1:12
Muchang Bahng
562
edited Nov 18 at 1:42
Robson
751221
edited Nov 18 at 1:42
Robson
751221
edited Nov 18 at 1:42
Robson
751221
751221
answered Nov 18 at 1:12
Muchang Bahng
562
answered Nov 18 at 1:12
Muchang Bahng
562
answered Nov 18 at 1:12
Muchang Bahng
562
562
1
Yeah I figured it out thank you!!!
– Minh Nguyen Nhat
Nov 18 at 1:15
1
hello @Muchang Bahng, you can use MathJax to improve your answers and help more people in a more beatiful way here is a link that could be useful for you math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 18 at 1:29
add a comment |
1
Yeah I figured it out thank you!!!
– Minh Nguyen Nhat
Nov 18 at 1:15
1
hello @Muchang Bahng, you can use MathJax to improve your answers and help more people in a more beatiful way here is a link that could be useful for you math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 18 at 1:29
1
1
Yeah I figured it out thank you!!!
– Minh Nguyen Nhat
Nov 18 at 1:15
Yeah I figured it out thank you!!!
– Minh Nguyen Nhat
Nov 18 at 1:15
1
1
hello @Muchang Bahng, you can use MathJax to improve your answers and help more people in a more beatiful way here is a link that could be useful for you math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 18 at 1:29
hello @Muchang Bahng, you can use MathJax to improve your answers and help more people in a more beatiful way here is a link that could be useful for you math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 18 at 1:29
add a comment |
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Sorry this is the first time I'm using this so I had no idea how to properly put an equation so there is an image
– Minh Nguyen Nhat
Nov 18 at 0:55
Hint: logarithms turn products into sums.
– NickD
Nov 18 at 1:01
Yeah I figured that part and got log10(1000!) - log10(999!) but don't know how to derive that one without using wolfram alpha
– Minh Nguyen Nhat
Nov 18 at 1:03
Keep it as a product: there's lots of cancellations.
– NickD
Nov 18 at 1:05
1
Ahh, I see what you mean! Thanks very much!
– Minh Nguyen Nhat
Nov 18 at 1:09