Mixing
The inequality $sum_{k=1}^n frac{1}{k^4} le 2 - frac{1}{sqrt n}$
Prove that for every $n$ we have $$sum_{k=1}^n frac{1}{k^4} le 2 - dfrac{1}{sqrt{n}}$$ I've tried induction, but I ended up with polynomials of high degree.
inequality summation
asked Nov 16 '18 at 11:29
J. Abraham
453314
add a comment |
Prove that for every $n$ we have $$sum_{k=1}^n frac{1}{k^4} le 2 - dfrac{1}{sqrt{n}}$$ I've tried induction, but I ended up with polynomials of high degree.
inequality summation
asked Nov 16 '18 at 11:29
J. Abraham
453314
Wrong for $n=2,3,4,5$ !
– Yves Daoust
Nov 16 '18 at 11:35
Presumably the inequality should go the other way around, since the limit as $n to infty$ of the l.h.s. is $< 2$.
– Travis
Nov 16 '18 at 12:00
Fixed, sorry for my mistake
– J. Abraham
Nov 16 '18 at 12:18
You should fix the name of you post, too ...
– Stockfish
Nov 16 '18 at 12:28
add a comment |
Prove that for every $n$ we have $$sum_{k=1}^n frac{1}{k^4} le 2 - dfrac{1}{sqrt{n}}$$ I've tried induction, but I ended up with polynomials of high degree.
inequality summation
asked Nov 16 '18 at 11:29
J. Abraham
453314
Prove that for every $n$ we have $$sum_{k=1}^n frac{1}{k^4} le 2 - dfrac{1}{sqrt{n}}$$ I've tried induction, but I ended up with polynomials of high degree.
inequality summation
inequality summation
asked Nov 16 '18 at 11:29
J. Abraham
453314
asked Nov 16 '18 at 11:29
J. Abraham
453314
edited Nov 16 '18 at 12:29
asked Nov 16 '18 at 11:29
J. Abraham
453314
asked Nov 16 '18 at 11:29
J. Abraham
453314
asked Nov 16 '18 at 11:29
J. Abraham
453314
453314
Wrong for $n=2,3,4,5$ !
– Yves Daoust
Nov 16 '18 at 11:35
Presumably the inequality should go the other way around, since the limit as $n to infty$ of the l.h.s. is $< 2$.
– Travis
Nov 16 '18 at 12:00
Fixed, sorry for my mistake
– J. Abraham
Nov 16 '18 at 12:18
You should fix the name of you post, too ...
– Stockfish
Nov 16 '18 at 12:28
add a comment |
Wrong for $n=2,3,4,5$ !
– Yves Daoust
Nov 16 '18 at 11:35
Presumably the inequality should go the other way around, since the limit as $n to infty$ of the l.h.s. is $< 2$.
– Travis
Nov 16 '18 at 12:00
Fixed, sorry for my mistake
– J. Abraham
Nov 16 '18 at 12:18
You should fix the name of you post, too ...
– Stockfish
Nov 16 '18 at 12:28
Wrong for $n=2,3,4,5$ !
– Yves Daoust
Nov 16 '18 at 11:35
Wrong for $n=2,3,4,5$ !
– Yves Daoust
Nov 16 '18 at 11:35
Presumably the inequality should go the other way around, since the limit as $n to infty$ of the l.h.s. is $< 2$.
– Travis
Nov 16 '18 at 12:00
Presumably the inequality should go the other way around, since the limit as $n to infty$ of the l.h.s. is $< 2$.
– Travis
Nov 16 '18 at 12:00
Fixed, sorry for my mistake
– J. Abraham
Nov 16 '18 at 12:18
Fixed, sorry for my mistake
– J. Abraham
Nov 16 '18 at 12:18
You should fix the name of you post, too ...
– Stockfish
Nov 16 '18 at 12:28
You should fix the name of you post, too ...
– Stockfish
Nov 16 '18 at 12:28
add a comment |
2 Answers
2
active
oldest
votes
For $nge2$,
$$
begin{align}
frac1{n^4}
&lefrac1{2n^{3/2}}\
&=frac1{sqrt{nvphantom{-1}}sqrt{nvphantom{-1}}left(sqrt{nvphantom{-1}}+sqrt{nvphantom{-1}}right)}\
&lefrac1{sqrt{nvphantom{-1}}sqrt{n-1}left(sqrt{nvphantom{-1}}+sqrt{n-1}right)}\
&=frac1{sqrt{n-1}}-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
Therefore,
$$
begin{align}
sum_{k=1}^nfrac1{k^4}
&le1+sum_{k=2}^nleft(frac1{sqrt{k-1}}-frac1{sqrt{kvphantom{-1}}}right)\
&=2-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
Note that this exact same estimate also works for $sumlimits_{k=1}^nfrac1{k^{5/2}}$
– robjohn♦
Nov 21 '18 at 17:47
add a comment |
The function $x^{-p}$ is a positive decreasing function. For such functions, sums at evenly spaced points are well approximated by integrals. More precisely,$$ int_1^n frac{1}{x^p}dx < sum_{i = 1}^n frac{1}{i^p} < int_1^n
frac{1}{x^p}dx + 1.$$
$$sum_{k=1}^n frac{1}{k^4}<int_1^n
frac{1}{x^4}dx + 1 =1+frac{n^{-3}}{-3}=1-frac{1}{3n^3}$$
Also, given that the function $$f(x)=3x^3sqrt{x}+sqrt{x}-3x^3$$
is strictly increasing and defined on $[0,+infty)$ with $f(0)=0$, we have
$$f(n) geq 0 iff 3n^3sqrt{n}+sqrt{n}-3n^3geq0 Rightarrow frac{3n^3sqrt{n}+sqrt{n}-3n^3}{3n^3sqrt{n}}> 0 iff$$
$$ 1+frac{1}{3n^3}-frac{1}{sqrt{n}} >0 Rightarrow1-frac{1}{3n^3}< 2-frac{1}{sqrt{n}}$$
So
$$sum_{k=1}^n frac{1}{k^4}leq2-frac{1}{sqrt{n}}$$
with equality only for $n=1$.
answered Nov 16 '18 at 12:31
Jevaut
788111
Can you solve it without calculus?
– J. Abraham
Nov 16 '18 at 12:44
For the first part, no. I can't think of a more efficient way than an integral bound. For the second part: $$ 1-frac{1}{3n^3} < 2 - frac{1}{sqrt{n}} iff -frac{1}{3n^3} < frac{sqrt{n}-1}{sqrt{n}} iff -frac{1}{3n^2sqrt{n}} < sqrt{n}-1$$ which is true, because $sqrt{n} geq 1$ (LHS is negative, RHS is non-negative).
– Jevaut
Nov 16 '18 at 13:00
add a comment |
Your Answer
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
For $nge2$,
$$
begin{align}
frac1{n^4}
&lefrac1{2n^{3/2}}\
&=frac1{sqrt{nvphantom{-1}}sqrt{nvphantom{-1}}left(sqrt{nvphantom{-1}}+sqrt{nvphantom{-1}}right)}\
&lefrac1{sqrt{nvphantom{-1}}sqrt{n-1}left(sqrt{nvphantom{-1}}+sqrt{n-1}right)}\
&=frac1{sqrt{n-1}}-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
Therefore,
$$
begin{align}
sum_{k=1}^nfrac1{k^4}
&le1+sum_{k=2}^nleft(frac1{sqrt{k-1}}-frac1{sqrt{kvphantom{-1}}}right)\
&=2-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
Note that this exact same estimate also works for $sumlimits_{k=1}^nfrac1{k^{5/2}}$
– robjohn♦
Nov 21 '18 at 17:47
add a comment |
For $nge2$,
$$
begin{align}
frac1{n^4}
&lefrac1{2n^{3/2}}\
&=frac1{sqrt{nvphantom{-1}}sqrt{nvphantom{-1}}left(sqrt{nvphantom{-1}}+sqrt{nvphantom{-1}}right)}\
&lefrac1{sqrt{nvphantom{-1}}sqrt{n-1}left(sqrt{nvphantom{-1}}+sqrt{n-1}right)}\
&=frac1{sqrt{n-1}}-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
Therefore,
$$
begin{align}
sum_{k=1}^nfrac1{k^4}
&le1+sum_{k=2}^nleft(frac1{sqrt{k-1}}-frac1{sqrt{kvphantom{-1}}}right)\
&=2-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
Note that this exact same estimate also works for $sumlimits_{k=1}^nfrac1{k^{5/2}}$
– robjohn♦
Nov 21 '18 at 17:47
add a comment |
For $nge2$,
$$
begin{align}
frac1{n^4}
&lefrac1{2n^{3/2}}\
&=frac1{sqrt{nvphantom{-1}}sqrt{nvphantom{-1}}left(sqrt{nvphantom{-1}}+sqrt{nvphantom{-1}}right)}\
&lefrac1{sqrt{nvphantom{-1}}sqrt{n-1}left(sqrt{nvphantom{-1}}+sqrt{n-1}right)}\
&=frac1{sqrt{n-1}}-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
Therefore,
$$
begin{align}
sum_{k=1}^nfrac1{k^4}
&le1+sum_{k=2}^nleft(frac1{sqrt{k-1}}-frac1{sqrt{kvphantom{-1}}}right)\
&=2-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
For $nge2$,
$$
begin{align}
frac1{n^4}
&lefrac1{2n^{3/2}}\
&=frac1{sqrt{nvphantom{-1}}sqrt{nvphantom{-1}}left(sqrt{nvphantom{-1}}+sqrt{nvphantom{-1}}right)}\
&lefrac1{sqrt{nvphantom{-1}}sqrt{n-1}left(sqrt{nvphantom{-1}}+sqrt{n-1}right)}\
&=frac1{sqrt{n-1}}-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
Therefore,
$$
begin{align}
sum_{k=1}^nfrac1{k^4}
&le1+sum_{k=2}^nleft(frac1{sqrt{k-1}}-frac1{sqrt{kvphantom{-1}}}right)\
&=2-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
answered Nov 21 '18 at 13:43
robjohn♦
265k27303624
265k27303624
Note that this exact same estimate also works for $sumlimits_{k=1}^nfrac1{k^{5/2}}$
– robjohn♦
Nov 21 '18 at 17:47
add a comment |
Note that this exact same estimate also works for $sumlimits_{k=1}^nfrac1{k^{5/2}}$
– robjohn♦
Nov 21 '18 at 17:47
Note that this exact same estimate also works for $sumlimits_{k=1}^nfrac1{k^{5/2}}$
– robjohn♦
Nov 21 '18 at 17:47
Note that this exact same estimate also works for $sumlimits_{k=1}^nfrac1{k^{5/2}}$
– robjohn♦
Nov 21 '18 at 17:47
add a comment |
The function $x^{-p}$ is a positive decreasing function. For such functions, sums at evenly spaced points are well approximated by integrals. More precisely,$$ int_1^n frac{1}{x^p}dx < sum_{i = 1}^n frac{1}{i^p} < int_1^n
frac{1}{x^p}dx + 1.$$
$$sum_{k=1}^n frac{1}{k^4}<int_1^n
frac{1}{x^4}dx + 1 =1+frac{n^{-3}}{-3}=1-frac{1}{3n^3}$$
Also, given that the function $$f(x)=3x^3sqrt{x}+sqrt{x}-3x^3$$
is strictly increasing and defined on $[0,+infty)$ with $f(0)=0$, we have
$$f(n) geq 0 iff 3n^3sqrt{n}+sqrt{n}-3n^3geq0 Rightarrow frac{3n^3sqrt{n}+sqrt{n}-3n^3}{3n^3sqrt{n}}> 0 iff$$
$$ 1+frac{1}{3n^3}-frac{1}{sqrt{n}} >0 Rightarrow1-frac{1}{3n^3}< 2-frac{1}{sqrt{n}}$$
So
$$sum_{k=1}^n frac{1}{k^4}leq2-frac{1}{sqrt{n}}$$
with equality only for $n=1$.
answered Nov 16 '18 at 12:31
Jevaut
788111
Can you solve it without calculus?
– J. Abraham
Nov 16 '18 at 12:44
For the first part, no. I can't think of a more efficient way than an integral bound. For the second part: $$ 1-frac{1}{3n^3} < 2 - frac{1}{sqrt{n}} iff -frac{1}{3n^3} < frac{sqrt{n}-1}{sqrt{n}} iff -frac{1}{3n^2sqrt{n}} < sqrt{n}-1$$ which is true, because $sqrt{n} geq 1$ (LHS is negative, RHS is non-negative).
– Jevaut
Nov 16 '18 at 13:00
add a comment |
The function $x^{-p}$ is a positive decreasing function. For such functions, sums at evenly spaced points are well approximated by integrals. More precisely,$$ int_1^n frac{1}{x^p}dx < sum_{i = 1}^n frac{1}{i^p} < int_1^n
frac{1}{x^p}dx + 1.$$
$$sum_{k=1}^n frac{1}{k^4}<int_1^n
frac{1}{x^4}dx + 1 =1+frac{n^{-3}}{-3}=1-frac{1}{3n^3}$$
Also, given that the function $$f(x)=3x^3sqrt{x}+sqrt{x}-3x^3$$
is strictly increasing and defined on $[0,+infty)$ with $f(0)=0$, we have
$$f(n) geq 0 iff 3n^3sqrt{n}+sqrt{n}-3n^3geq0 Rightarrow frac{3n^3sqrt{n}+sqrt{n}-3n^3}{3n^3sqrt{n}}> 0 iff$$
$$ 1+frac{1}{3n^3}-frac{1}{sqrt{n}} >0 Rightarrow1-frac{1}{3n^3}< 2-frac{1}{sqrt{n}}$$
So
$$sum_{k=1}^n frac{1}{k^4}leq2-frac{1}{sqrt{n}}$$
with equality only for $n=1$.
answered Nov 16 '18 at 12:31
Jevaut
788111
Can you solve it without calculus?
– J. Abraham
Nov 16 '18 at 12:44
For the first part, no. I can't think of a more efficient way than an integral bound. For the second part: $$ 1-frac{1}{3n^3} < 2 - frac{1}{sqrt{n}} iff -frac{1}{3n^3} < frac{sqrt{n}-1}{sqrt{n}} iff -frac{1}{3n^2sqrt{n}} < sqrt{n}-1$$ which is true, because $sqrt{n} geq 1$ (LHS is negative, RHS is non-negative).
– Jevaut
Nov 16 '18 at 13:00
add a comment |
The function $x^{-p}$ is a positive decreasing function. For such functions, sums at evenly spaced points are well approximated by integrals. More precisely,$$ int_1^n frac{1}{x^p}dx < sum_{i = 1}^n frac{1}{i^p} < int_1^n
frac{1}{x^p}dx + 1.$$
$$sum_{k=1}^n frac{1}{k^4}<int_1^n
frac{1}{x^4}dx + 1 =1+frac{n^{-3}}{-3}=1-frac{1}{3n^3}$$
Also, given that the function $$f(x)=3x^3sqrt{x}+sqrt{x}-3x^3$$
is strictly increasing and defined on $[0,+infty)$ with $f(0)=0$, we have
$$f(n) geq 0 iff 3n^3sqrt{n}+sqrt{n}-3n^3geq0 Rightarrow frac{3n^3sqrt{n}+sqrt{n}-3n^3}{3n^3sqrt{n}}> 0 iff$$
$$ 1+frac{1}{3n^3}-frac{1}{sqrt{n}} >0 Rightarrow1-frac{1}{3n^3}< 2-frac{1}{sqrt{n}}$$
So
$$sum_{k=1}^n frac{1}{k^4}leq2-frac{1}{sqrt{n}}$$
with equality only for $n=1$.
answered Nov 16 '18 at 12:31
Jevaut
788111
The function $x^{-p}$ is a positive decreasing function. For such functions, sums at evenly spaced points are well approximated by integrals. More precisely,$$ int_1^n frac{1}{x^p}dx < sum_{i = 1}^n frac{1}{i^p} < int_1^n
frac{1}{x^p}dx + 1.$$
$$sum_{k=1}^n frac{1}{k^4}<int_1^n
frac{1}{x^4}dx + 1 =1+frac{n^{-3}}{-3}=1-frac{1}{3n^3}$$
Also, given that the function $$f(x)=3x^3sqrt{x}+sqrt{x}-3x^3$$
is strictly increasing and defined on $[0,+infty)$ with $f(0)=0$, we have
$$f(n) geq 0 iff 3n^3sqrt{n}+sqrt{n}-3n^3geq0 Rightarrow frac{3n^3sqrt{n}+sqrt{n}-3n^3}{3n^3sqrt{n}}> 0 iff$$
$$ 1+frac{1}{3n^3}-frac{1}{sqrt{n}} >0 Rightarrow1-frac{1}{3n^3}< 2-frac{1}{sqrt{n}}$$
So
$$sum_{k=1}^n frac{1}{k^4}leq2-frac{1}{sqrt{n}}$$
with equality only for $n=1$.
answered Nov 16 '18 at 12:31
Jevaut
788111
edited Nov 21 '18 at 6:41
answered Nov 16 '18 at 12:31
Jevaut
788111
answered Nov 16 '18 at 12:31
Jevaut
788111
answered Nov 16 '18 at 12:31
Jevaut
788111
788111
Can you solve it without calculus?
– J. Abraham
Nov 16 '18 at 12:44
For the first part, no. I can't think of a more efficient way than an integral bound. For the second part: $$ 1-frac{1}{3n^3} < 2 - frac{1}{sqrt{n}} iff -frac{1}{3n^3} < frac{sqrt{n}-1}{sqrt{n}} iff -frac{1}{3n^2sqrt{n}} < sqrt{n}-1$$ which is true, because $sqrt{n} geq 1$ (LHS is negative, RHS is non-negative).
– Jevaut
Nov 16 '18 at 13:00
add a comment |
Can you solve it without calculus?
– J. Abraham
Nov 16 '18 at 12:44
For the first part, no. I can't think of a more efficient way than an integral bound. For the second part: $$ 1-frac{1}{3n^3} < 2 - frac{1}{sqrt{n}} iff -frac{1}{3n^3} < frac{sqrt{n}-1}{sqrt{n}} iff -frac{1}{3n^2sqrt{n}} < sqrt{n}-1$$ which is true, because $sqrt{n} geq 1$ (LHS is negative, RHS is non-negative).
– Jevaut
Nov 16 '18 at 13:00
Can you solve it without calculus?
– J. Abraham
Nov 16 '18 at 12:44
Can you solve it without calculus?
– J. Abraham
Nov 16 '18 at 12:44
For the first part, no. I can't think of a more efficient way than an integral bound. For the second part: $$ 1-frac{1}{3n^3} < 2 - frac{1}{sqrt{n}} iff -frac{1}{3n^3} < frac{sqrt{n}-1}{sqrt{n}} iff -frac{1}{3n^2sqrt{n}} < sqrt{n}-1$$ which is true, because $sqrt{n} geq 1$ (LHS is negative, RHS is non-negative).
– Jevaut
Nov 16 '18 at 13:00
For the first part, no. I can't think of a more efficient way than an integral bound. For the second part: $$ 1-frac{1}{3n^3} < 2 - frac{1}{sqrt{n}} iff -frac{1}{3n^3} < frac{sqrt{n}-1}{sqrt{n}} iff -frac{1}{3n^2sqrt{n}} < sqrt{n}-1$$ which is true, because $sqrt{n} geq 1$ (LHS is negative, RHS is non-negative).
– Jevaut
Nov 16 '18 at 13:00
add a comment |
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Wrong for $n=2,3,4,5$ !
– Yves Daoust
Nov 16 '18 at 11:35
Presumably the inequality should go the other way around, since the limit as $n to infty$ of the l.h.s. is $< 2$.
– Travis
Nov 16 '18 at 12:00
Fixed, sorry for my mistake
– J. Abraham
Nov 16 '18 at 12:18
You should fix the name of you post, too ...
– Stockfish
Nov 16 '18 at 12:28