Prove $sum_i=1^nfrac1sqrtilefrac1sqrtn!prod_i=2^n(sqrti-1)+2sum_i=2^nfrac1sqrti$ using Weierstrass inequalityProving the Power Mean Inequality using Chebyshev's sum inequalityHow prove this inequality $fracasqrtb^2+b+1+fracbsqrtc^2+c+1+fraccsqrta^2+a+1gesqrt3$Generalized Bernoulli's inequality of the form $frac11-sum_i=1^nx_igeqprod_i=1^n(1+x_i)$How prove Reversing the Arithmetic mean – Geometric mean inequality？Prove that $ sqrtfracab+3 +sqrtfracbc+3 +sqrtfracca+3 leq frac32 $How to prove this inequality? ThanksProve inequality using Jensen's inequalityInequality : $ sum_cyc fraca+bsqrta+2c geq 2 sqrta+b+c$How to prove the Weierstrass product inequality using recurrency?
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Prove $sum_i=1^nfrac1sqrtilefrac1sqrtn!prod_i=2^n(sqrti-1)+2sum_i=2^nfrac1sqrti$ using Weierstrass inequality
Proving the Power Mean Inequality using Chebyshev's sum inequalityHow prove this inequality $fracasqrtb^2+b+1+fracbsqrtc^2+c+1+fraccsqrta^2+a+1gesqrt3$Generalized Bernoulli's inequality of the form $frac11-sum_i=1^nx_igeqprod_i=1^n(1+x_i)$How prove Reversing the Arithmetic mean – Geometric mean inequality？Prove that $ sqrtfracab+3 +sqrtfracbc+3 +sqrtfracca+3 leq frac32 $How to prove this inequality? ThanksProve inequality using Jensen's inequalityInequality : $ sum_cyc fraca+bsqrta+2c geq 2 sqrta+b+c$How to prove the Weierstrass product inequality using recurrency?
$begingroup$
Use Weierstrass's inequality to prove that
$$sum_i=1^n frac1sqrtile frac1sqrtn! prod_i = 2^n (sqrti-1) + 2 left(sum_i = 2^n frac1sqrtiright).$$
Using Weierstrass's inequality, I get
$$1 + sum_i=1^n frac1sqrti le frac1sqrtn! left(prod_i = 1^n (1 + sqrti)right)$$
real-analysis inequality
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
asked Mar 21 at 17:29
JekiJeki
53
$endgroup$
add a comment |
$begingroup$
Use Weierstrass's inequality to prove that
$$sum_i=1^n frac1sqrtile frac1sqrtn! prod_i = 2^n (sqrti-1) + 2 left(sum_i = 2^n frac1sqrtiright).$$
Using Weierstrass's inequality, I get
$$1 + sum_i=1^n frac1sqrti le frac1sqrtn! left(prod_i = 1^n (1 + sqrti)right)$$
real-analysis inequality
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
asked Mar 21 at 17:29
JekiJeki
53
$endgroup$
$begingroup$
The very same question has already been asked here, so please use approach0 to search before you ask, to see if your question (likely a homework problem if two people ask it in such a short time frame) has been ask before.
$endgroup$
– Viktor Glombik
Mar 21 at 18:09
add a comment |
$begingroup$
Use Weierstrass's inequality to prove that
$$sum_i=1^n frac1sqrtile frac1sqrtn! prod_i = 2^n (sqrti-1) + 2 left(sum_i = 2^n frac1sqrtiright).$$
Using Weierstrass's inequality, I get
$$1 + sum_i=1^n frac1sqrti le frac1sqrtn! left(prod_i = 1^n (1 + sqrti)right)$$
real-analysis inequality
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
asked Mar 21 at 17:29
JekiJeki
53
$endgroup$
Use Weierstrass's inequality to prove that
$$sum_i=1^n frac1sqrtile frac1sqrtn! prod_i = 2^n (sqrti-1) + 2 left(sum_i = 2^n frac1sqrtiright).$$
Using Weierstrass's inequality, I get
$$1 + sum_i=1^n frac1sqrti le frac1sqrtn! left(prod_i = 1^n (1 + sqrti)right)$$
real-analysis inequality
real-analysis inequality
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
asked Mar 21 at 17:29
JekiJeki
53
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
asked Mar 21 at 17:29
JekiJeki
53
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
edited Mar 21 at 18:04
Viktor Glombik
1,3122628
1,3122628
asked Mar 21 at 17:29
JekiJeki
53
asked Mar 21 at 17:29
JekiJeki
53
asked Mar 21 at 17:29
JekiJeki
53
53
$begingroup$
The very same question has already been asked here, so please use approach0 to search before you ask, to see if your question (likely a homework problem if two people ask it in such a short time frame) has been ask before.
$endgroup$
– Viktor Glombik
Mar 21 at 18:09
add a comment |
$begingroup$
The very same question has already been asked here, so please use approach0 to search before you ask, to see if your question (likely a homework problem if two people ask it in such a short time frame) has been ask before.
$endgroup$
– Viktor Glombik
Mar 21 at 18:09
$begingroup$
The very same question has already been asked here, so please use approach0 to search before you ask, to see if your question (likely a homework problem if two people ask it in such a short time frame) has been ask before.
$endgroup$
– Viktor Glombik
Mar 21 at 18:09
$begingroup$
The very same question has already been asked here, so please use approach0 to search before you ask, to see if your question (likely a homework problem if two people ask it in such a short time frame) has been ask before.
$endgroup$
– Viktor Glombik
Mar 21 at 18:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Since
$$
sum_k = 1^n frac1sqrtk = 1 + sum_k = 2^n frac1sqrtk,
$$
we can rewrite the inequality, which has to be shown (by subtracting $2 sum_k = 2^n frac1sqrtk$ from both sides) as
$$
1 - sum_k = 2^n frac1sqrtk
le frac1sqrtn! prod_k = 2^n (sqrtk-1).
$$
By the Weierstrass inequality (since $frac1sqrtk in (0,1]$ for all $k in mathbbN_>0$) we have
$$
1 - sum_k = 2^n frac1sqrtk
le prod_k = 2^n left(1 - frac1sqrtkright)
= prod_k = 2^n fracsqrtk - 1sqrtk
= frac1sqrtn! prod_k = 2^n (sqrtk-1),
le frac1sqrtn! prod_k = 2^n sqrtk - 1
$$
where the last equality is achieved by pulling out the denominator out of the product and the last inequality is using the fact that $sqrtk-1 le sqrtk - 1$ for all $k in mathbbN_>1$, which you can easily verify either by induction or by drawing a graph.
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Since
$$
sum_k = 1^n frac1sqrtk = 1 + sum_k = 2^n frac1sqrtk,
$$
we can rewrite the inequality, which has to be shown (by subtracting $2 sum_k = 2^n frac1sqrtk$ from both sides) as
$$
1 - sum_k = 2^n frac1sqrtk
le frac1sqrtn! prod_k = 2^n (sqrtk-1).
$$
By the Weierstrass inequality (since $frac1sqrtk in (0,1]$ for all $k in mathbbN_>0$) we have
$$
1 - sum_k = 2^n frac1sqrtk
le prod_k = 2^n left(1 - frac1sqrtkright)
= prod_k = 2^n fracsqrtk - 1sqrtk
= frac1sqrtn! prod_k = 2^n (sqrtk-1),
le frac1sqrtn! prod_k = 2^n sqrtk - 1
$$
where the last equality is achieved by pulling out the denominator out of the product and the last inequality is using the fact that $sqrtk-1 le sqrtk - 1$ for all $k in mathbbN_>1$, which you can easily verify either by induction or by drawing a graph.
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
$endgroup$
add a comment |
$begingroup$
Hint:
Since
$$
sum_k = 1^n frac1sqrtk = 1 + sum_k = 2^n frac1sqrtk,
$$
we can rewrite the inequality, which has to be shown (by subtracting $2 sum_k = 2^n frac1sqrtk$ from both sides) as
$$
1 - sum_k = 2^n frac1sqrtk
le frac1sqrtn! prod_k = 2^n (sqrtk-1).
$$
By the Weierstrass inequality (since $frac1sqrtk in (0,1]$ for all $k in mathbbN_>0$) we have
$$
1 - sum_k = 2^n frac1sqrtk
le prod_k = 2^n left(1 - frac1sqrtkright)
= prod_k = 2^n fracsqrtk - 1sqrtk
= frac1sqrtn! prod_k = 2^n (sqrtk-1),
le frac1sqrtn! prod_k = 2^n sqrtk - 1
$$
where the last equality is achieved by pulling out the denominator out of the product and the last inequality is using the fact that $sqrtk-1 le sqrtk - 1$ for all $k in mathbbN_>1$, which you can easily verify either by induction or by drawing a graph.
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
$endgroup$
add a comment |
$begingroup$
Hint:
Since
$$
sum_k = 1^n frac1sqrtk = 1 + sum_k = 2^n frac1sqrtk,
$$
we can rewrite the inequality, which has to be shown (by subtracting $2 sum_k = 2^n frac1sqrtk$ from both sides) as
$$
1 - sum_k = 2^n frac1sqrtk
le frac1sqrtn! prod_k = 2^n (sqrtk-1).
$$
By the Weierstrass inequality (since $frac1sqrtk in (0,1]$ for all $k in mathbbN_>0$) we have
$$
1 - sum_k = 2^n frac1sqrtk
le prod_k = 2^n left(1 - frac1sqrtkright)
= prod_k = 2^n fracsqrtk - 1sqrtk
= frac1sqrtn! prod_k = 2^n (sqrtk-1),
le frac1sqrtn! prod_k = 2^n sqrtk - 1
$$
where the last equality is achieved by pulling out the denominator out of the product and the last inequality is using the fact that $sqrtk-1 le sqrtk - 1$ for all $k in mathbbN_>1$, which you can easily verify either by induction or by drawing a graph.
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
$endgroup$
Hint:
Since
$$
sum_k = 1^n frac1sqrtk = 1 + sum_k = 2^n frac1sqrtk,
$$
we can rewrite the inequality, which has to be shown (by subtracting $2 sum_k = 2^n frac1sqrtk$ from both sides) as
$$
1 - sum_k = 2^n frac1sqrtk
le frac1sqrtn! prod_k = 2^n (sqrtk-1).
$$
By the Weierstrass inequality (since $frac1sqrtk in (0,1]$ for all $k in mathbbN_>0$) we have
$$
1 - sum_k = 2^n frac1sqrtk
le prod_k = 2^n left(1 - frac1sqrtkright)
= prod_k = 2^n fracsqrtk - 1sqrtk
= frac1sqrtn! prod_k = 2^n (sqrtk-1),
le frac1sqrtn! prod_k = 2^n sqrtk - 1
$$
where the last equality is achieved by pulling out the denominator out of the product and the last inequality is using the fact that $sqrtk-1 le sqrtk - 1$ for all $k in mathbbN_>1$, which you can easily verify either by induction or by drawing a graph.
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
edited Mar 21 at 17:57
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
answered Mar 21 at 17:49
Viktor GlombikViktor Glombik
1,3122628
1,3122628
add a comment |
add a comment |
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$begingroup$
The very same question has already been asked here, so please use approach0 to search before you ask, to see if your question (likely a homework problem if two people ask it in such a short time frame) has been ask before.
$endgroup$
– Viktor Glombik
Mar 21 at 18:09