Compute $limlimits_n to infty frac1 cdot 3 cdot 5 cdots(2n - 1)2 cdot 4 cdot 6 cdots (2n)$ [duplicate]
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This question already has an answer here:
Evaluate $prod frac 2k - 1 2k$ [duplicate]
2 answers
Convergence point of $fracprod_k=1^n (2k-1) prod_k=1^n 2k$
4 answers
EDIT: @Holo has kindly pointed out that my concept of ln rules used in this question is wrong. However, the intuition behind using the tangent of a curve to find the sum to infinity of a series still stands. Therefore, I won't be editing the post.
I tried expanding out the equation from the question and got $$a_n = frac12 .frac34 . frac56 ...frac2n -12n$$
I then tried taking the ln of the equation which works out to $$ln(1 - frac12) + ln(1 - frac14) + ...$$
Here is my question. I used the rules of ln functions and took $ln(frac11/2) + ln(frac1frac14)$ + ...
Since ln(1) is 0, shouldn't the limit work out to 0 as n tends to infinity? Then $$ln(L) = 0, where L = limit$$
$$L = e^0$$
$$L = 1$$
However, the limit is actually 0, and my tutor used a method which I couldn't understand as he tried approximating the function $y = ln(x)$ to $y = x - 1$, as he says that the linear equation is actually a tangent to the ln curve. Could someone please explain this intuition behind it? He differentiated the equation, and since the curve cuts the x-axis at x = 1, he got the linear curve $y = x - 1$.
He did mention that the linear equation is just an approximation, but said such an approximation would be more than sufficient.
calculus limits
edited Sep 6 at 6:03
David G. Stork
8,15621232
asked Sep 6 at 5:11
statsguy21
353
marked as duplicate by JavaMan, Nosrati, A. Pongrácz, Lord Shark the Unknown, Jose Arnaldo Bebita Dris Sep 6 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 4 more comments
up vote
1
down vote
favorite
This question already has an answer here:
Evaluate $prod frac 2k - 1 2k$ [duplicate]
2 answers
Convergence point of $fracprod_k=1^n (2k-1) prod_k=1^n 2k$
4 answers
EDIT: @Holo has kindly pointed out that my concept of ln rules used in this question is wrong. However, the intuition behind using the tangent of a curve to find the sum to infinity of a series still stands. Therefore, I won't be editing the post.
I tried expanding out the equation from the question and got $$a_n = frac12 .frac34 . frac56 ...frac2n -12n$$
I then tried taking the ln of the equation which works out to $$ln(1 - frac12) + ln(1 - frac14) + ...$$
Here is my question. I used the rules of ln functions and took $ln(frac11/2) + ln(frac1frac14)$ + ...
Since ln(1) is 0, shouldn't the limit work out to 0 as n tends to infinity? Then $$ln(L) = 0, where L = limit$$
$$L = e^0$$
$$L = 1$$
However, the limit is actually 0, and my tutor used a method which I couldn't understand as he tried approximating the function $y = ln(x)$ to $y = x - 1$, as he says that the linear equation is actually a tangent to the ln curve. Could someone please explain this intuition behind it? He differentiated the equation, and since the curve cuts the x-axis at x = 1, he got the linear curve $y = x - 1$.
He did mention that the linear equation is just an approximation, but said such an approximation would be more than sufficient.
calculus limits
edited Sep 6 at 6:03
David G. Stork
8,15621232
asked Sep 6 at 5:11
statsguy21
353
marked as duplicate by JavaMan, Nosrati, A. Pongrácz, Lord Shark the Unknown, Jose Arnaldo Bebita Dris Sep 6 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
How did you got to $ln(frac11/2) + ln(frac1frac14)$, this is wrong, the rule says that $ln(a)-ln(b)=ln(a/b)$ not that $ln(a-b)=ln(a/b)$
– Holo
Sep 6 at 5:18
I see...so my concept was totally wrong in the first place. Should I edit it or should I just keep it up?
– statsguy21
Sep 6 at 5:23
Please look for similar questions before posting your new question. math.stackexchange.com/questions/2886753/…, math.stackexchange.com/questions/1586773/…, math.stackexchange.com/questions/93001/…, math.stackexchange.com/questions/2402508/…
– JavaMan
Sep 6 at 5:26
@statsguy keep it up, or if edit just explain you made a mistake. It is better to show effort in the post(something that you clearly did)
– Holo
Sep 6 at 5:28
@JavaMan I don't think this is a duplicate, this is question about intuition behind the linear approximation, and not the product itself(Or so I think)
– Holo
Sep 6 at 5:28
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Evaluate $prod frac 2k - 1 2k$ [duplicate]
2 answers
Convergence point of $fracprod_k=1^n (2k-1) prod_k=1^n 2k$
4 answers
EDIT: @Holo has kindly pointed out that my concept of ln rules used in this question is wrong. However, the intuition behind using the tangent of a curve to find the sum to infinity of a series still stands. Therefore, I won't be editing the post.
I tried expanding out the equation from the question and got $$a_n = frac12 .frac34 . frac56 ...frac2n -12n$$
I then tried taking the ln of the equation which works out to $$ln(1 - frac12) + ln(1 - frac14) + ...$$
Here is my question. I used the rules of ln functions and took $ln(frac11/2) + ln(frac1frac14)$ + ...
Since ln(1) is 0, shouldn't the limit work out to 0 as n tends to infinity? Then $$ln(L) = 0, where L = limit$$
$$L = e^0$$
$$L = 1$$
However, the limit is actually 0, and my tutor used a method which I couldn't understand as he tried approximating the function $y = ln(x)$ to $y = x - 1$, as he says that the linear equation is actually a tangent to the ln curve. Could someone please explain this intuition behind it? He differentiated the equation, and since the curve cuts the x-axis at x = 1, he got the linear curve $y = x - 1$.
He did mention that the linear equation is just an approximation, but said such an approximation would be more than sufficient.
calculus limits
edited Sep 6 at 6:03
David G. Stork
8,15621232
asked Sep 6 at 5:11
statsguy21
353
This question already has an answer here:
Evaluate $prod frac 2k - 1 2k$ [duplicate]
2 answers
Convergence point of $fracprod_k=1^n (2k-1) prod_k=1^n 2k$
4 answers
EDIT: @Holo has kindly pointed out that my concept of ln rules used in this question is wrong. However, the intuition behind using the tangent of a curve to find the sum to infinity of a series still stands. Therefore, I won't be editing the post.
I tried expanding out the equation from the question and got $$a_n = frac12 .frac34 . frac56 ...frac2n -12n$$
I then tried taking the ln of the equation which works out to $$ln(1 - frac12) + ln(1 - frac14) + ...$$
Here is my question. I used the rules of ln functions and took $ln(frac11/2) + ln(frac1frac14)$ + ...
Since ln(1) is 0, shouldn't the limit work out to 0 as n tends to infinity? Then $$ln(L) = 0, where L = limit$$
$$L = e^0$$
$$L = 1$$
However, the limit is actually 0, and my tutor used a method which I couldn't understand as he tried approximating the function $y = ln(x)$ to $y = x - 1$, as he says that the linear equation is actually a tangent to the ln curve. Could someone please explain this intuition behind it? He differentiated the equation, and since the curve cuts the x-axis at x = 1, he got the linear curve $y = x - 1$.
He did mention that the linear equation is just an approximation, but said such an approximation would be more than sufficient.
This question already has an answer here:
Evaluate $prod frac 2k - 1 2k$ [duplicate]
2 answers
Convergence point of $fracprod_k=1^n (2k-1) prod_k=1^n 2k$
4 answers
calculus limits
calculus limits
edited Sep 6 at 6:03
David G. Stork
8,15621232
asked Sep 6 at 5:11
statsguy21
353
edited Sep 6 at 6:03
David G. Stork
8,15621232
asked Sep 6 at 5:11
statsguy21
353
edited Sep 6 at 6:03
David G. Stork
8,15621232
edited Sep 6 at 6:03
David G. Stork
8,15621232
edited Sep 6 at 6:03
David G. Stork
8,15621232
8,15621232
asked Sep 6 at 5:11
statsguy21
353
asked Sep 6 at 5:11
statsguy21
353
asked Sep 6 at 5:11
statsguy21
353
353
marked as duplicate by JavaMan, Nosrati, A. Pongrácz, Lord Shark the Unknown, Jose Arnaldo Bebita Dris Sep 6 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by JavaMan, Nosrati, A. Pongrácz, Lord Shark the Unknown, Jose Arnaldo Bebita Dris Sep 6 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
How did you got to $ln(frac11/2) + ln(frac1frac14)$, this is wrong, the rule says that $ln(a)-ln(b)=ln(a/b)$ not that $ln(a-b)=ln(a/b)$
– Holo
Sep 6 at 5:18
I see...so my concept was totally wrong in the first place. Should I edit it or should I just keep it up?
– statsguy21
Sep 6 at 5:23
Please look for similar questions before posting your new question. math.stackexchange.com/questions/2886753/…, math.stackexchange.com/questions/1586773/…, math.stackexchange.com/questions/93001/…, math.stackexchange.com/questions/2402508/…
– JavaMan
Sep 6 at 5:26
@statsguy keep it up, or if edit just explain you made a mistake. It is better to show effort in the post(something that you clearly did)
– Holo
Sep 6 at 5:28
@JavaMan I don't think this is a duplicate, this is question about intuition behind the linear approximation, and not the product itself(Or so I think)
– Holo
Sep 6 at 5:28
 |Â
show 4 more comments
1
How did you got to $ln(frac11/2) + ln(frac1frac14)$, this is wrong, the rule says that $ln(a)-ln(b)=ln(a/b)$ not that $ln(a-b)=ln(a/b)$
– Holo
Sep 6 at 5:18
I see...so my concept was totally wrong in the first place. Should I edit it or should I just keep it up?
– statsguy21
Sep 6 at 5:23
Please look for similar questions before posting your new question. math.stackexchange.com/questions/2886753/…, math.stackexchange.com/questions/1586773/…, math.stackexchange.com/questions/93001/…, math.stackexchange.com/questions/2402508/…
– JavaMan
Sep 6 at 5:26
@statsguy keep it up, or if edit just explain you made a mistake. It is better to show effort in the post(something that you clearly did)
– Holo
Sep 6 at 5:28
@JavaMan I don't think this is a duplicate, this is question about intuition behind the linear approximation, and not the product itself(Or so I think)
– Holo
Sep 6 at 5:28
1
1
How did you got to $ln(frac11/2) + ln(frac1frac14)$, this is wrong, the rule says that $ln(a)-ln(b)=ln(a/b)$ not that $ln(a-b)=ln(a/b)$
– Holo
Sep 6 at 5:18
How did you got to $ln(frac11/2) + ln(frac1frac14)$, this is wrong, the rule says that $ln(a)-ln(b)=ln(a/b)$ not that $ln(a-b)=ln(a/b)$
– Holo
Sep 6 at 5:18
I see...so my concept was totally wrong in the first place. Should I edit it or should I just keep it up?
– statsguy21
Sep 6 at 5:23
I see...so my concept was totally wrong in the first place. Should I edit it or should I just keep it up?
– statsguy21
Sep 6 at 5:23
Please look for similar questions before posting your new question. math.stackexchange.com/questions/2886753/…, math.stackexchange.com/questions/1586773/…, math.stackexchange.com/questions/93001/…, math.stackexchange.com/questions/2402508/…
– JavaMan
Sep 6 at 5:26
Please look for similar questions before posting your new question. math.stackexchange.com/questions/2886753/…, math.stackexchange.com/questions/1586773/…, math.stackexchange.com/questions/93001/…, math.stackexchange.com/questions/2402508/…
– JavaMan
Sep 6 at 5:26
@statsguy keep it up, or if edit just explain you made a mistake. It is better to show effort in the post(something that you clearly did)
– Holo
Sep 6 at 5:28
@statsguy keep it up, or if edit just explain you made a mistake. It is better to show effort in the post(something that you clearly did)
– Holo
Sep 6 at 5:28
@JavaMan I don't think this is a duplicate, this is question about intuition behind the linear approximation, and not the product itself(Or so I think)
– Holo
Sep 6 at 5:28
@JavaMan I don't think this is a duplicate, this is question about intuition behind the linear approximation, and not the product itself(Or so I think)
– Holo
Sep 6 at 5:28
 |Â
show 4 more comments
3 Answers
3
active
oldest
votes
up vote
6
down vote
A completely different approach is to write
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n=frac (2n)!(2^nn!)^2$$
because you can divide a $2$ out of each term on the bottom and get $n!$ and you can multiply top and bottom by the bottom. Now feed it to Stirling
$$a_napproxfrac(2n)^2ne^2ne^2n2^2nn^2nsqrtpi n=frac 1sqrtpi nto 0$$
answered Sep 6 at 5:22
Ross Millikan
281k23191358
nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background)
– dineshdileep
Sep 6 at 5:40
1
@dineshdileep That's not true in general. Example $prod _2^infty (1-2/(n(n+1))) = 1/3$.
– xbh
Sep 6 at 5:45
@dineshdileep No, you can't say that. A well-known counterexample is $lim_n to infty(1-frac1n)^n = frac1e$
– ab123
Sep 6 at 5:46
Powerful as always.
– mrs
Sep 6 at 5:53
add a comment |Â
up vote
1
down vote
As you say,
$$ln a_n=sum_k=1^nlnleft(1-frac12kright).$$
But as $xto0$,
$$ln(1-x)=-x+O(x^2).$$
Therefore
$$ln a_n=-sum_k=1^nfrac12k+Oleft(sum_k=1^nfrac1k^2right).$$
As the series $sum_1^infty1/k$ diverges and $sum_1^infty1/k^2$ converges,
then $ln a_nto-infty$, and so $a_nto0$.
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
add a comment |Â
up vote
1
down vote
We can prove that
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n<frac1sqrt2n+1.$$
Let $b_n=frac23 .frac45 . frac67 ...frac2n2n+1$, then it is easy to see that $a_n<b_n$. So $$a_n^2<a_nb_n=frac12n+1.$$
You limit is $$lim_nto inftyfrac12 .frac34 . frac56 ...frac2n -12n=0.$$
answered Sep 6 at 6:00
Riemann
3,0881321
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
A completely different approach is to write
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n=frac (2n)!(2^nn!)^2$$
because you can divide a $2$ out of each term on the bottom and get $n!$ and you can multiply top and bottom by the bottom. Now feed it to Stirling
$$a_napproxfrac(2n)^2ne^2ne^2n2^2nn^2nsqrtpi n=frac 1sqrtpi nto 0$$
answered Sep 6 at 5:22
Ross Millikan
281k23191358
nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background)
– dineshdileep
Sep 6 at 5:40
1
@dineshdileep That's not true in general. Example $prod _2^infty (1-2/(n(n+1))) = 1/3$.
– xbh
Sep 6 at 5:45
@dineshdileep No, you can't say that. A well-known counterexample is $lim_n to infty(1-frac1n)^n = frac1e$
– ab123
Sep 6 at 5:46
Powerful as always.
– mrs
Sep 6 at 5:53
add a comment |Â
up vote
6
down vote
A completely different approach is to write
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n=frac (2n)!(2^nn!)^2$$
because you can divide a $2$ out of each term on the bottom and get $n!$ and you can multiply top and bottom by the bottom. Now feed it to Stirling
$$a_napproxfrac(2n)^2ne^2ne^2n2^2nn^2nsqrtpi n=frac 1sqrtpi nto 0$$
answered Sep 6 at 5:22
Ross Millikan
281k23191358
nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background)
– dineshdileep
Sep 6 at 5:40
1
@dineshdileep That's not true in general. Example $prod _2^infty (1-2/(n(n+1))) = 1/3$.
– xbh
Sep 6 at 5:45
@dineshdileep No, you can't say that. A well-known counterexample is $lim_n to infty(1-frac1n)^n = frac1e$
– ab123
Sep 6 at 5:46
Powerful as always.
– mrs
Sep 6 at 5:53
add a comment |Â
up vote
6
down vote
up vote
6
down vote
A completely different approach is to write
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n=frac (2n)!(2^nn!)^2$$
because you can divide a $2$ out of each term on the bottom and get $n!$ and you can multiply top and bottom by the bottom. Now feed it to Stirling
$$a_napproxfrac(2n)^2ne^2ne^2n2^2nn^2nsqrtpi n=frac 1sqrtpi nto 0$$
answered Sep 6 at 5:22
Ross Millikan
281k23191358
A completely different approach is to write
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n=frac (2n)!(2^nn!)^2$$
because you can divide a $2$ out of each term on the bottom and get $n!$ and you can multiply top and bottom by the bottom. Now feed it to Stirling
$$a_napproxfrac(2n)^2ne^2ne^2n2^2nn^2nsqrtpi n=frac 1sqrtpi nto 0$$
answered Sep 6 at 5:22
Ross Millikan
281k23191358
answered Sep 6 at 5:22
Ross Millikan
281k23191358
answered Sep 6 at 5:22
Ross Millikan
281k23191358
answered Sep 6 at 5:22
Ross Millikan
281k23191358
281k23191358
nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background)
– dineshdileep
Sep 6 at 5:40
1
@dineshdileep That's not true in general. Example $prod _2^infty (1-2/(n(n+1))) = 1/3$.
– xbh
Sep 6 at 5:45
@dineshdileep No, you can't say that. A well-known counterexample is $lim_n to infty(1-frac1n)^n = frac1e$
– ab123
Sep 6 at 5:46
Powerful as always.
– mrs
Sep 6 at 5:53
add a comment |Â
nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background)
– dineshdileep
Sep 6 at 5:40
1
@dineshdileep That's not true in general. Example $prod _2^infty (1-2/(n(n+1))) = 1/3$.
– xbh
Sep 6 at 5:45
@dineshdileep No, you can't say that. A well-known counterexample is $lim_n to infty(1-frac1n)^n = frac1e$
– ab123
Sep 6 at 5:46
Powerful as always.
– mrs
Sep 6 at 5:53
nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background)
– dineshdileep
Sep 6 at 5:40
nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background)
– dineshdileep
Sep 6 at 5:40
1
1
@dineshdileep That's not true in general. Example $prod _2^infty (1-2/(n(n+1))) = 1/3$.
– xbh
Sep 6 at 5:45
@dineshdileep That's not true in general. Example $prod _2^infty (1-2/(n(n+1))) = 1/3$.
– xbh
Sep 6 at 5:45
@dineshdileep No, you can't say that. A well-known counterexample is $lim_n to infty(1-frac1n)^n = frac1e$
– ab123
Sep 6 at 5:46
@dineshdileep No, you can't say that. A well-known counterexample is $lim_n to infty(1-frac1n)^n = frac1e$
– ab123
Sep 6 at 5:46
Powerful as always.
– mrs
Sep 6 at 5:53
Powerful as always.
– mrs
Sep 6 at 5:53
add a comment |Â
up vote
1
down vote
As you say,
$$ln a_n=sum_k=1^nlnleft(1-frac12kright).$$
But as $xto0$,
$$ln(1-x)=-x+O(x^2).$$
Therefore
$$ln a_n=-sum_k=1^nfrac12k+Oleft(sum_k=1^nfrac1k^2right).$$
As the series $sum_1^infty1/k$ diverges and $sum_1^infty1/k^2$ converges,
then $ln a_nto-infty$, and so $a_nto0$.
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
add a comment |Â
up vote
1
down vote
As you say,
$$ln a_n=sum_k=1^nlnleft(1-frac12kright).$$
But as $xto0$,
$$ln(1-x)=-x+O(x^2).$$
Therefore
$$ln a_n=-sum_k=1^nfrac12k+Oleft(sum_k=1^nfrac1k^2right).$$
As the series $sum_1^infty1/k$ diverges and $sum_1^infty1/k^2$ converges,
then $ln a_nto-infty$, and so $a_nto0$.
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As you say,
$$ln a_n=sum_k=1^nlnleft(1-frac12kright).$$
But as $xto0$,
$$ln(1-x)=-x+O(x^2).$$
Therefore
$$ln a_n=-sum_k=1^nfrac12k+Oleft(sum_k=1^nfrac1k^2right).$$
As the series $sum_1^infty1/k$ diverges and $sum_1^infty1/k^2$ converges,
then $ln a_nto-infty$, and so $a_nto0$.
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
As you say,
$$ln a_n=sum_k=1^nlnleft(1-frac12kright).$$
But as $xto0$,
$$ln(1-x)=-x+O(x^2).$$
Therefore
$$ln a_n=-sum_k=1^nfrac12k+Oleft(sum_k=1^nfrac1k^2right).$$
As the series $sum_1^infty1/k$ diverges and $sum_1^infty1/k^2$ converges,
then $ln a_nto-infty$, and so $a_nto0$.
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
answered Sep 6 at 5:40
Lord Shark the Unknown
89.8k955117
89.8k955117
add a comment |Â
add a comment |Â
up vote
1
down vote
We can prove that
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n<frac1sqrt2n+1.$$
Let $b_n=frac23 .frac45 . frac67 ...frac2n2n+1$, then it is easy to see that $a_n<b_n$. So $$a_n^2<a_nb_n=frac12n+1.$$
You limit is $$lim_nto inftyfrac12 .frac34 . frac56 ...frac2n -12n=0.$$
answered Sep 6 at 6:00
Riemann
3,0881321
add a comment |Â
up vote
1
down vote
We can prove that
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n<frac1sqrt2n+1.$$
Let $b_n=frac23 .frac45 . frac67 ...frac2n2n+1$, then it is easy to see that $a_n<b_n$. So $$a_n^2<a_nb_n=frac12n+1.$$
You limit is $$lim_nto inftyfrac12 .frac34 . frac56 ...frac2n -12n=0.$$
answered Sep 6 at 6:00
Riemann
3,0881321
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We can prove that
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n<frac1sqrt2n+1.$$
Let $b_n=frac23 .frac45 . frac67 ...frac2n2n+1$, then it is easy to see that $a_n<b_n$. So $$a_n^2<a_nb_n=frac12n+1.$$
You limit is $$lim_nto inftyfrac12 .frac34 . frac56 ...frac2n -12n=0.$$
answered Sep 6 at 6:00
Riemann
3,0881321
We can prove that
$$a_n = frac12 .frac34 . frac56 ...frac2n -12n<frac1sqrt2n+1.$$
Let $b_n=frac23 .frac45 . frac67 ...frac2n2n+1$, then it is easy to see that $a_n<b_n$. So $$a_n^2<a_nb_n=frac12n+1.$$
You limit is $$lim_nto inftyfrac12 .frac34 . frac56 ...frac2n -12n=0.$$
answered Sep 6 at 6:00
Riemann
3,0881321
answered Sep 6 at 6:00
Riemann
3,0881321
answered Sep 6 at 6:00
Riemann
3,0881321
answered Sep 6 at 6:00
Riemann
3,0881321
3,0881321
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1
How did you got to $ln(frac11/2) + ln(frac1frac14)$, this is wrong, the rule says that $ln(a)-ln(b)=ln(a/b)$ not that $ln(a-b)=ln(a/b)$
– Holo
Sep 6 at 5:18
I see...so my concept was totally wrong in the first place. Should I edit it or should I just keep it up?
– statsguy21
Sep 6 at 5:23
Please look for similar questions before posting your new question. math.stackexchange.com/questions/2886753/…, math.stackexchange.com/questions/1586773/…, math.stackexchange.com/questions/93001/…, math.stackexchange.com/questions/2402508/…
– JavaMan
Sep 6 at 5:26
@statsguy keep it up, or if edit just explain you made a mistake. It is better to show effort in the post(something that you clearly did)
– Holo
Sep 6 at 5:28
@JavaMan I don't think this is a duplicate, this is question about intuition behind the linear approximation, and not the product itself(Or so I think)
– Holo
Sep 6 at 5:28