Evaluating $lim_{ntoinfty}sqrt[n]frac{prod_{k=1}^{n}(2n+k)}{n^n}$
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How to find $$lim_{ntoinfty} sqrt[n]frac{prod_{k=1}^{n}(2n+k)}{n^n}quad ?$$
Wolfram Alpha says that it's $frac{27}{4{e}}$, but I fail to arrive at this result.
Methods allowed in my class so far are pretty limited - I've tried treating the product in the numerator as a polynomial (it's a product of n linear terms) and finding the coefficient at $n^n$, the highest power - but it seems to be the wrong approach. May anyone help me, please?
calculus real-analysis limits
edited Nov 16 at 14:50
StubbornAtom
4,86911137
asked Nov 16 at 3:14
Bag of Chips
864
add a comment |
up vote
0
down vote
favorite
How to find $$lim_{ntoinfty} sqrt[n]frac{prod_{k=1}^{n}(2n+k)}{n^n}quad ?$$
Wolfram Alpha says that it's $frac{27}{4{e}}$, but I fail to arrive at this result.
Methods allowed in my class so far are pretty limited - I've tried treating the product in the numerator as a polynomial (it's a product of n linear terms) and finding the coefficient at $n^n$, the highest power - but it seems to be the wrong approach. May anyone help me, please?
calculus real-analysis limits
edited Nov 16 at 14:50
StubbornAtom
4,86911137
asked Nov 16 at 3:14
Bag of Chips
864
1
Looks like a Riemann sum if you take logarithm of $$prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}$$
– StubbornAtom
Nov 16 at 3:22
Sorry, take logarithm of $$left[prod_{k=1}^nleft( frac{2n+k}{n}right)right]^{1/n} $$ and then take the limit.
– StubbornAtom
Nov 16 at 3:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to find $$lim_{ntoinfty} sqrt[n]frac{prod_{k=1}^{n}(2n+k)}{n^n}quad ?$$
Wolfram Alpha says that it's $frac{27}{4{e}}$, but I fail to arrive at this result.
Methods allowed in my class so far are pretty limited - I've tried treating the product in the numerator as a polynomial (it's a product of n linear terms) and finding the coefficient at $n^n$, the highest power - but it seems to be the wrong approach. May anyone help me, please?
calculus real-analysis limits
edited Nov 16 at 14:50
StubbornAtom
4,86911137
asked Nov 16 at 3:14
Bag of Chips
864
How to find $$lim_{ntoinfty} sqrt[n]frac{prod_{k=1}^{n}(2n+k)}{n^n}quad ?$$
Wolfram Alpha says that it's $frac{27}{4{e}}$, but I fail to arrive at this result.
Methods allowed in my class so far are pretty limited - I've tried treating the product in the numerator as a polynomial (it's a product of n linear terms) and finding the coefficient at $n^n$, the highest power - but it seems to be the wrong approach. May anyone help me, please?
calculus real-analysis limits
calculus real-analysis limits
edited Nov 16 at 14:50
StubbornAtom
4,86911137
asked Nov 16 at 3:14
Bag of Chips
864
edited Nov 16 at 14:50
StubbornAtom
4,86911137
asked Nov 16 at 3:14
Bag of Chips
864
edited Nov 16 at 14:50
StubbornAtom
4,86911137
edited Nov 16 at 14:50
StubbornAtom
4,86911137
edited Nov 16 at 14:50
StubbornAtom
4,86911137
4,86911137
asked Nov 16 at 3:14
Bag of Chips
864
asked Nov 16 at 3:14
Bag of Chips
864
asked Nov 16 at 3:14
Bag of Chips
864
864
1
Looks like a Riemann sum if you take logarithm of $$prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}$$
– StubbornAtom
Nov 16 at 3:22
Sorry, take logarithm of $$left[prod_{k=1}^nleft( frac{2n+k}{n}right)right]^{1/n} $$ and then take the limit.
– StubbornAtom
Nov 16 at 3:30
add a comment |
1
Looks like a Riemann sum if you take logarithm of $$prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}$$
– StubbornAtom
Nov 16 at 3:22
Sorry, take logarithm of $$left[prod_{k=1}^nleft( frac{2n+k}{n}right)right]^{1/n} $$ and then take the limit.
– StubbornAtom
Nov 16 at 3:30
1
1
Looks like a Riemann sum if you take logarithm of $$prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}$$
– StubbornAtom
Nov 16 at 3:22
Looks like a Riemann sum if you take logarithm of $$prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}$$
– StubbornAtom
Nov 16 at 3:22
Sorry, take logarithm of $$left[prod_{k=1}^nleft( frac{2n+k}{n}right)right]^{1/n} $$ and then take the limit.
– StubbornAtom
Nov 16 at 3:30
Sorry, take logarithm of $$left[prod_{k=1}^nleft( frac{2n+k}{n}right)right]^{1/n} $$ and then take the limit.
– StubbornAtom
Nov 16 at 3:30
add a comment |
1 Answer
1
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oldest
votes
up vote
5
down vote
Not too bad after taking the log,
$$logleft(lim_{ntoinfty}prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}right) = lim_{ntoinfty}sum_{k=1}^nfrac{1}{n}logleft(2+tfrac{k}{n}right) = int_0^1log(2+x)dx$$
Should be able to do the rest from here.
answered Nov 16 at 3:32
Nicholas Parris
1182
Of course, order of taking log and limit does not matter.
– StubbornAtom
Nov 16 at 14:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Not too bad after taking the log,
$$logleft(lim_{ntoinfty}prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}right) = lim_{ntoinfty}sum_{k=1}^nfrac{1}{n}logleft(2+tfrac{k}{n}right) = int_0^1log(2+x)dx$$
Should be able to do the rest from here.
answered Nov 16 at 3:32
Nicholas Parris
1182
Of course, order of taking log and limit does not matter.
– StubbornAtom
Nov 16 at 14:48
add a comment |
up vote
5
down vote
Not too bad after taking the log,
$$logleft(lim_{ntoinfty}prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}right) = lim_{ntoinfty}sum_{k=1}^nfrac{1}{n}logleft(2+tfrac{k}{n}right) = int_0^1log(2+x)dx$$
Should be able to do the rest from here.
answered Nov 16 at 3:32
Nicholas Parris
1182
Of course, order of taking log and limit does not matter.
– StubbornAtom
Nov 16 at 14:48
add a comment |
up vote
5
down vote
up vote
5
down vote
Not too bad after taking the log,
$$logleft(lim_{ntoinfty}prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}right) = lim_{ntoinfty}sum_{k=1}^nfrac{1}{n}logleft(2+tfrac{k}{n}right) = int_0^1log(2+x)dx$$
Should be able to do the rest from here.
answered Nov 16 at 3:32
Nicholas Parris
1182
Not too bad after taking the log,
$$logleft(lim_{ntoinfty}prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}right) = lim_{ntoinfty}sum_{k=1}^nfrac{1}{n}logleft(2+tfrac{k}{n}right) = int_0^1log(2+x)dx$$
Should be able to do the rest from here.
answered Nov 16 at 3:32
Nicholas Parris
1182
answered Nov 16 at 3:32
Nicholas Parris
1182
answered Nov 16 at 3:32
Nicholas Parris
1182
answered Nov 16 at 3:32
Nicholas Parris
1182
1182
Of course, order of taking log and limit does not matter.
– StubbornAtom
Nov 16 at 14:48
add a comment |
Of course, order of taking log and limit does not matter.
– StubbornAtom
Nov 16 at 14:48
Of course, order of taking log and limit does not matter.
– StubbornAtom
Nov 16 at 14:48
Of course, order of taking log and limit does not matter.
– StubbornAtom
Nov 16 at 14:48
add a comment |
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1
Looks like a Riemann sum if you take logarithm of $$prod_{k=1}^nleft( frac{2n+k}{n}right)^{1/n}$$
– StubbornAtom
Nov 16 at 3:22
Sorry, take logarithm of $$left[prod_{k=1}^nleft( frac{2n+k}{n}right)right]^{1/n} $$ and then take the limit.
– StubbornAtom
Nov 16 at 3:30