Evaluating $lim_{ntoinfty}sqrt[n]frac{prod_{k=1}^{n}(2n+k)}{n^n}$
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
4,89911137
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
4,89911137
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
4,89911137
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
4,89911137
asked Nov 16 at 3:14
Bag of Chips
864
edited Nov 16 at 14:50
StubbornAtom
4,89911137
asked Nov 16 at 3:14
Bag of Chips
864
edited Nov 16 at 14:50
StubbornAtom
4,89911137
edited Nov 16 at 14:50
StubbornAtom
4,89911137
edited Nov 16 at 14:50
StubbornAtom
4,89911137
4,89911137
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
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
1183
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
1183
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
1183
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
1183
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
1183
answered Nov 16 at 3:32
Nicholas Parris
1183
answered Nov 16 at 3:32
Nicholas Parris
1183
answered Nov 16 at 3:32
Nicholas Parris
1183
1183
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000668%2fevaluating-lim-n-to-infty-sqrtn-frac-prod-k-1n2nknn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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