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For $a> 0$, determine $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)$?
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For $a> 0$, determine $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)$
My attempts : $sum_k=1^n( a^frackn^2 - 1) = ( a^frac1n^2- 1) + ( a^frac2n^2- 1)+......... +(a^fracnn^2- 1)=a^frac1n^2+ a^frac2n^2++.....+a^fracnn^2- n $
ThereFore $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= a^lim_nrightarrow inftyfrac1n^2+ a^lim_nrightarrow inftyfrac2n^2++.....+a^lim_nrightarrow inftyfracnn^2 - lim_nrightarrow infty n = lim_nrightarrow infty (a^0 + a^0 +.....+a^0) -lim_nrightarrow infty n = lim_nrightarrow infty( 1+1.....+1) -lim_nrightarrow infty n=lim_nrightarrow infty n -lim_nrightarrow infty n = 0 $
Hence $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= 0$
Is my answer is correct ??
Any hints / solution will be appreciated
thanks u..
real-analysis
asked Jul 31 at 23:26
stupid
52418
add a comment |Â
up vote
3
down vote
favorite
For $a> 0$, determine $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)$
My attempts : $sum_k=1^n( a^frackn^2 - 1) = ( a^frac1n^2- 1) + ( a^frac2n^2- 1)+......... +(a^fracnn^2- 1)=a^frac1n^2+ a^frac2n^2++.....+a^fracnn^2- n $
ThereFore $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= a^lim_nrightarrow inftyfrac1n^2+ a^lim_nrightarrow inftyfrac2n^2++.....+a^lim_nrightarrow inftyfracnn^2 - lim_nrightarrow infty n = lim_nrightarrow infty (a^0 + a^0 +.....+a^0) -lim_nrightarrow infty n = lim_nrightarrow infty( 1+1.....+1) -lim_nrightarrow infty n=lim_nrightarrow infty n -lim_nrightarrow infty n = 0 $
Hence $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= 0$
Is my answer is correct ??
Any hints / solution will be appreciated
thanks u..
real-analysis
asked Jul 31 at 23:26
stupid
52418
4
Your answer is not correct. You cannot bring the limit into the exponent the way you have because there are precisely $n$ copies of $a$. By moving the limit to the exponent, you’ve made the number of $a$’s constant.
– Clayton
Jul 31 at 23:55
2
Indeterminate form $infty - infty$ requires more study, you cannot simply say $0$.
– GEdgar
Aug 1 at 0:48
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For $a> 0$, determine $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)$
My attempts : $sum_k=1^n( a^frackn^2 - 1) = ( a^frac1n^2- 1) + ( a^frac2n^2- 1)+......... +(a^fracnn^2- 1)=a^frac1n^2+ a^frac2n^2++.....+a^fracnn^2- n $
ThereFore $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= a^lim_nrightarrow inftyfrac1n^2+ a^lim_nrightarrow inftyfrac2n^2++.....+a^lim_nrightarrow inftyfracnn^2 - lim_nrightarrow infty n = lim_nrightarrow infty (a^0 + a^0 +.....+a^0) -lim_nrightarrow infty n = lim_nrightarrow infty( 1+1.....+1) -lim_nrightarrow infty n=lim_nrightarrow infty n -lim_nrightarrow infty n = 0 $
Hence $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= 0$
Is my answer is correct ??
Any hints / solution will be appreciated
thanks u..
real-analysis
asked Jul 31 at 23:26
stupid
52418
For $a> 0$, determine $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)$
My attempts : $sum_k=1^n( a^frackn^2 - 1) = ( a^frac1n^2- 1) + ( a^frac2n^2- 1)+......... +(a^fracnn^2- 1)=a^frac1n^2+ a^frac2n^2++.....+a^fracnn^2- n $
ThereFore $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= a^lim_nrightarrow inftyfrac1n^2+ a^lim_nrightarrow inftyfrac2n^2++.....+a^lim_nrightarrow inftyfracnn^2 - lim_nrightarrow infty n = lim_nrightarrow infty (a^0 + a^0 +.....+a^0) -lim_nrightarrow infty n = lim_nrightarrow infty( 1+1.....+1) -lim_nrightarrow infty n=lim_nrightarrow infty n -lim_nrightarrow infty n = 0 $
Hence $lim_nrightarrow inftysum_k=1^n( a^frackn^2 - 1)= 0$
Is my answer is correct ??
Any hints / solution will be appreciated
thanks u..
real-analysis
asked Jul 31 at 23:26
stupid
52418
asked Jul 31 at 23:26
stupid
52418
asked Jul 31 at 23:26
stupid
52418
asked Jul 31 at 23:26
stupid
52418
52418
4
Your answer is not correct. You cannot bring the limit into the exponent the way you have because there are precisely $n$ copies of $a$. By moving the limit to the exponent, you’ve made the number of $a$’s constant.
– Clayton
Jul 31 at 23:55
2
Indeterminate form $infty - infty$ requires more study, you cannot simply say $0$.
– GEdgar
Aug 1 at 0:48
add a comment |Â
4
Your answer is not correct. You cannot bring the limit into the exponent the way you have because there are precisely $n$ copies of $a$. By moving the limit to the exponent, you’ve made the number of $a$’s constant.
– Clayton
Jul 31 at 23:55
2
Indeterminate form $infty - infty$ requires more study, you cannot simply say $0$.
– GEdgar
Aug 1 at 0:48
4
4
Your answer is not correct. You cannot bring the limit into the exponent the way you have because there are precisely $n$ copies of $a$. By moving the limit to the exponent, you’ve made the number of $a$’s constant.
– Clayton
Jul 31 at 23:55
Your answer is not correct. You cannot bring the limit into the exponent the way you have because there are precisely $n$ copies of $a$. By moving the limit to the exponent, you’ve made the number of $a$’s constant.
– Clayton
Jul 31 at 23:55
2
2
Indeterminate form $infty - infty$ requires more study, you cannot simply say $0$.
– GEdgar
Aug 1 at 0:48
Indeterminate form $infty - infty$ requires more study, you cannot simply say $0$.
– GEdgar
Aug 1 at 0:48
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Intuition:
since $e^u = 1+u + o(u)$ when $uto 0$ and
$0 leq frackn^2ln a leq fracln an xrightarrow[ntoinfty] 0$ for every $1leq kleq n$, we "expect" to have
$$
sum_k=1^n( a^frackn^2-1)
= sum_k=1^n( e^frackn^2ln a-1) approx sum_k=1^n frackn^2ln a = ln acdot fracn(n+1)2n^2xrightarrow[ntoinfty] fracln a2 tag1
$$
so we should try and make this rigorous. (Sanity check: this is consistent with $a=1$, for which we do get $0$.)
Side note. Easy lower bound: using the standard inequality (proven e.g. by convexity) $e^x geq 1+x$ for all $x$, we have
$$
sum_k=1^n( e^frackn^2ln a-1) geq sum_k=1^nfrackn^2ln a = fracln a2cdot fracn(n+1)n^2 tag2
$$
and as mentioned we have $lim_ntoinfty fracln a2cdot fracn(n+1)n^2 = fracln a2$. So we have that the limit is at least $fracln a2$... That was the easy part.
Proof of (1).
We can write
$$beginalign
sum_k=1^n( e^frackn^2ln a-1)
&=sum_k=1^nunderbracebig(e^fracln an^2big) _=alpha^k - n = alphacdot frac1-alpha^n1-alpha - n
= e^fracln an^2frac1-e^fracln an1-e^fracln an^2 - n \
&= left(1+ oleft(frac1nright)right)cdot fracfracln an+fracln^2 a2n^2+o(frac1n^2)fracln an^2+o(frac1n^2) - n\
&= left(1+ oleft(frac1nright)right)cdot fracn+fracln a2+o(1)1+o(1) - n\
&= n+fracln a2+o(1) - n\
&= fracln a2+o(1) xrightarrow[ntoinfty] fracln a2 tag3
endalign$$
where we used the Taylor series of $e^u$ as $uto 0$, $e^u = 1+u+fracu^22+o(u)$, and that of $frac11+u = 1-u+o(u)$. This last (3) gives us the expected limit, $fracln a2$.
answered Aug 1 at 0:04
Clement C.
46.9k33682
beautiful solution and thanks u actually $a^frackn^2-1= e^frackn^2ln a-1$ that doesnot come in my minds......Now i got it... thanks u @Clement C.
– stupid
Aug 1 at 0:20
If $alpha - 1=b$ then the expression under limit is $((1+b)^n+1-1-b-nb)/b$ or $n(n+1)b/2+dots $ and this tends to $(log a) /2$.
– Paramanand Singh
Aug 1 at 7:49
@ParamanandSingh this is the limit I showed, yes :)
– Clement C.
Aug 1 at 13:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Intuition:
since $e^u = 1+u + o(u)$ when $uto 0$ and
$0 leq frackn^2ln a leq fracln an xrightarrow[ntoinfty] 0$ for every $1leq kleq n$, we "expect" to have
$$
sum_k=1^n( a^frackn^2-1)
= sum_k=1^n( e^frackn^2ln a-1) approx sum_k=1^n frackn^2ln a = ln acdot fracn(n+1)2n^2xrightarrow[ntoinfty] fracln a2 tag1
$$
so we should try and make this rigorous. (Sanity check: this is consistent with $a=1$, for which we do get $0$.)
Side note. Easy lower bound: using the standard inequality (proven e.g. by convexity) $e^x geq 1+x$ for all $x$, we have
$$
sum_k=1^n( e^frackn^2ln a-1) geq sum_k=1^nfrackn^2ln a = fracln a2cdot fracn(n+1)n^2 tag2
$$
and as mentioned we have $lim_ntoinfty fracln a2cdot fracn(n+1)n^2 = fracln a2$. So we have that the limit is at least $fracln a2$... That was the easy part.
Proof of (1).
We can write
$$beginalign
sum_k=1^n( e^frackn^2ln a-1)
&=sum_k=1^nunderbracebig(e^fracln an^2big) _=alpha^k - n = alphacdot frac1-alpha^n1-alpha - n
= e^fracln an^2frac1-e^fracln an1-e^fracln an^2 - n \
&= left(1+ oleft(frac1nright)right)cdot fracfracln an+fracln^2 a2n^2+o(frac1n^2)fracln an^2+o(frac1n^2) - n\
&= left(1+ oleft(frac1nright)right)cdot fracn+fracln a2+o(1)1+o(1) - n\
&= n+fracln a2+o(1) - n\
&= fracln a2+o(1) xrightarrow[ntoinfty] fracln a2 tag3
endalign$$
where we used the Taylor series of $e^u$ as $uto 0$, $e^u = 1+u+fracu^22+o(u)$, and that of $frac11+u = 1-u+o(u)$. This last (3) gives us the expected limit, $fracln a2$.
answered Aug 1 at 0:04
Clement C.
46.9k33682
beautiful solution and thanks u actually $a^frackn^2-1= e^frackn^2ln a-1$ that doesnot come in my minds......Now i got it... thanks u @Clement C.
– stupid
Aug 1 at 0:20
If $alpha - 1=b$ then the expression under limit is $((1+b)^n+1-1-b-nb)/b$ or $n(n+1)b/2+dots $ and this tends to $(log a) /2$.
– Paramanand Singh
Aug 1 at 7:49
@ParamanandSingh this is the limit I showed, yes :)
– Clement C.
Aug 1 at 13:37
add a comment |Â
up vote
3
down vote
accepted
Intuition:
since $e^u = 1+u + o(u)$ when $uto 0$ and
$0 leq frackn^2ln a leq fracln an xrightarrow[ntoinfty] 0$ for every $1leq kleq n$, we "expect" to have
$$
sum_k=1^n( a^frackn^2-1)
= sum_k=1^n( e^frackn^2ln a-1) approx sum_k=1^n frackn^2ln a = ln acdot fracn(n+1)2n^2xrightarrow[ntoinfty] fracln a2 tag1
$$
so we should try and make this rigorous. (Sanity check: this is consistent with $a=1$, for which we do get $0$.)
Side note. Easy lower bound: using the standard inequality (proven e.g. by convexity) $e^x geq 1+x$ for all $x$, we have
$$
sum_k=1^n( e^frackn^2ln a-1) geq sum_k=1^nfrackn^2ln a = fracln a2cdot fracn(n+1)n^2 tag2
$$
and as mentioned we have $lim_ntoinfty fracln a2cdot fracn(n+1)n^2 = fracln a2$. So we have that the limit is at least $fracln a2$... That was the easy part.
Proof of (1).
We can write
$$beginalign
sum_k=1^n( e^frackn^2ln a-1)
&=sum_k=1^nunderbracebig(e^fracln an^2big) _=alpha^k - n = alphacdot frac1-alpha^n1-alpha - n
= e^fracln an^2frac1-e^fracln an1-e^fracln an^2 - n \
&= left(1+ oleft(frac1nright)right)cdot fracfracln an+fracln^2 a2n^2+o(frac1n^2)fracln an^2+o(frac1n^2) - n\
&= left(1+ oleft(frac1nright)right)cdot fracn+fracln a2+o(1)1+o(1) - n\
&= n+fracln a2+o(1) - n\
&= fracln a2+o(1) xrightarrow[ntoinfty] fracln a2 tag3
endalign$$
where we used the Taylor series of $e^u$ as $uto 0$, $e^u = 1+u+fracu^22+o(u)$, and that of $frac11+u = 1-u+o(u)$. This last (3) gives us the expected limit, $fracln a2$.
answered Aug 1 at 0:04
Clement C.
46.9k33682
beautiful solution and thanks u actually $a^frackn^2-1= e^frackn^2ln a-1$ that doesnot come in my minds......Now i got it... thanks u @Clement C.
– stupid
Aug 1 at 0:20
If $alpha - 1=b$ then the expression under limit is $((1+b)^n+1-1-b-nb)/b$ or $n(n+1)b/2+dots $ and this tends to $(log a) /2$.
– Paramanand Singh
Aug 1 at 7:49
@ParamanandSingh this is the limit I showed, yes :)
– Clement C.
Aug 1 at 13:37
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Intuition:
since $e^u = 1+u + o(u)$ when $uto 0$ and
$0 leq frackn^2ln a leq fracln an xrightarrow[ntoinfty] 0$ for every $1leq kleq n$, we "expect" to have
$$
sum_k=1^n( a^frackn^2-1)
= sum_k=1^n( e^frackn^2ln a-1) approx sum_k=1^n frackn^2ln a = ln acdot fracn(n+1)2n^2xrightarrow[ntoinfty] fracln a2 tag1
$$
so we should try and make this rigorous. (Sanity check: this is consistent with $a=1$, for which we do get $0$.)
Side note. Easy lower bound: using the standard inequality (proven e.g. by convexity) $e^x geq 1+x$ for all $x$, we have
$$
sum_k=1^n( e^frackn^2ln a-1) geq sum_k=1^nfrackn^2ln a = fracln a2cdot fracn(n+1)n^2 tag2
$$
and as mentioned we have $lim_ntoinfty fracln a2cdot fracn(n+1)n^2 = fracln a2$. So we have that the limit is at least $fracln a2$... That was the easy part.
Proof of (1).
We can write
$$beginalign
sum_k=1^n( e^frackn^2ln a-1)
&=sum_k=1^nunderbracebig(e^fracln an^2big) _=alpha^k - n = alphacdot frac1-alpha^n1-alpha - n
= e^fracln an^2frac1-e^fracln an1-e^fracln an^2 - n \
&= left(1+ oleft(frac1nright)right)cdot fracfracln an+fracln^2 a2n^2+o(frac1n^2)fracln an^2+o(frac1n^2) - n\
&= left(1+ oleft(frac1nright)right)cdot fracn+fracln a2+o(1)1+o(1) - n\
&= n+fracln a2+o(1) - n\
&= fracln a2+o(1) xrightarrow[ntoinfty] fracln a2 tag3
endalign$$
where we used the Taylor series of $e^u$ as $uto 0$, $e^u = 1+u+fracu^22+o(u)$, and that of $frac11+u = 1-u+o(u)$. This last (3) gives us the expected limit, $fracln a2$.
answered Aug 1 at 0:04
Clement C.
46.9k33682
Intuition:
since $e^u = 1+u + o(u)$ when $uto 0$ and
$0 leq frackn^2ln a leq fracln an xrightarrow[ntoinfty] 0$ for every $1leq kleq n$, we "expect" to have
$$
sum_k=1^n( a^frackn^2-1)
= sum_k=1^n( e^frackn^2ln a-1) approx sum_k=1^n frackn^2ln a = ln acdot fracn(n+1)2n^2xrightarrow[ntoinfty] fracln a2 tag1
$$
so we should try and make this rigorous. (Sanity check: this is consistent with $a=1$, for which we do get $0$.)
Side note. Easy lower bound: using the standard inequality (proven e.g. by convexity) $e^x geq 1+x$ for all $x$, we have
$$
sum_k=1^n( e^frackn^2ln a-1) geq sum_k=1^nfrackn^2ln a = fracln a2cdot fracn(n+1)n^2 tag2
$$
and as mentioned we have $lim_ntoinfty fracln a2cdot fracn(n+1)n^2 = fracln a2$. So we have that the limit is at least $fracln a2$... That was the easy part.
Proof of (1).
We can write
$$beginalign
sum_k=1^n( e^frackn^2ln a-1)
&=sum_k=1^nunderbracebig(e^fracln an^2big) _=alpha^k - n = alphacdot frac1-alpha^n1-alpha - n
= e^fracln an^2frac1-e^fracln an1-e^fracln an^2 - n \
&= left(1+ oleft(frac1nright)right)cdot fracfracln an+fracln^2 a2n^2+o(frac1n^2)fracln an^2+o(frac1n^2) - n\
&= left(1+ oleft(frac1nright)right)cdot fracn+fracln a2+o(1)1+o(1) - n\
&= n+fracln a2+o(1) - n\
&= fracln a2+o(1) xrightarrow[ntoinfty] fracln a2 tag3
endalign$$
where we used the Taylor series of $e^u$ as $uto 0$, $e^u = 1+u+fracu^22+o(u)$, and that of $frac11+u = 1-u+o(u)$. This last (3) gives us the expected limit, $fracln a2$.
answered Aug 1 at 0:04
Clement C.
46.9k33682
edited Aug 1 at 0:11
answered Aug 1 at 0:04
Clement C.
46.9k33682
answered Aug 1 at 0:04
Clement C.
46.9k33682
answered Aug 1 at 0:04
Clement C.
46.9k33682
46.9k33682
beautiful solution and thanks u actually $a^frackn^2-1= e^frackn^2ln a-1$ that doesnot come in my minds......Now i got it... thanks u @Clement C.
– stupid
Aug 1 at 0:20
If $alpha - 1=b$ then the expression under limit is $((1+b)^n+1-1-b-nb)/b$ or $n(n+1)b/2+dots $ and this tends to $(log a) /2$.
– Paramanand Singh
Aug 1 at 7:49
@ParamanandSingh this is the limit I showed, yes :)
– Clement C.
Aug 1 at 13:37
add a comment |Â
beautiful solution and thanks u actually $a^frackn^2-1= e^frackn^2ln a-1$ that doesnot come in my minds......Now i got it... thanks u @Clement C.
– stupid
Aug 1 at 0:20
If $alpha - 1=b$ then the expression under limit is $((1+b)^n+1-1-b-nb)/b$ or $n(n+1)b/2+dots $ and this tends to $(log a) /2$.
– Paramanand Singh
Aug 1 at 7:49
@ParamanandSingh this is the limit I showed, yes :)
– Clement C.
Aug 1 at 13:37
beautiful solution and thanks u actually $a^frackn^2-1= e^frackn^2ln a-1$ that doesnot come in my minds......Now i got it... thanks u @Clement C.
– stupid
Aug 1 at 0:20
beautiful solution and thanks u actually $a^frackn^2-1= e^frackn^2ln a-1$ that doesnot come in my minds......Now i got it... thanks u @Clement C.
– stupid
Aug 1 at 0:20
If $alpha - 1=b$ then the expression under limit is $((1+b)^n+1-1-b-nb)/b$ or $n(n+1)b/2+dots $ and this tends to $(log a) /2$.
– Paramanand Singh
Aug 1 at 7:49
If $alpha - 1=b$ then the expression under limit is $((1+b)^n+1-1-b-nb)/b$ or $n(n+1)b/2+dots $ and this tends to $(log a) /2$.
– Paramanand Singh
Aug 1 at 7:49
@ParamanandSingh this is the limit I showed, yes :)
– Clement C.
Aug 1 at 13:37
@ParamanandSingh this is the limit I showed, yes :)
– Clement C.
Aug 1 at 13:37
add a comment |Â
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4
Your answer is not correct. You cannot bring the limit into the exponent the way you have because there are precisely $n$ copies of $a$. By moving the limit to the exponent, you’ve made the number of $a$’s constant.
– Clayton
Jul 31 at 23:55
2
Indeterminate form $infty - infty$ requires more study, you cannot simply say $0$.
– GEdgar
Aug 1 at 0:48