Is $1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x$












2














If $x ge 5$, is $1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x$



I believe the answer is yes.



Here is my thinking:



(1) $log_2{5} > 2.32 > 2.284 > 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5}$



(2) Assume up to $x$ that $log_2 x > sumlimits_{i le x}frac{1}{i}$



(3) $(x+1)^{x+1} > {{x+1}choose{x+1}}x^{x+1} + {{x+1}choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$



(4) $(x+1)log_2left(frac{x+1}{x}right) > 1$



(5) $log_2(x+1) - log_2(x) > frac{1}{x+1}$



(6) $log_2(x+1) - log_2(x) + log_2(x) = log_2(x+1) > sumlimits_{i le x+1}frac{1}{i}$






proof-verification inequality logarithms harmonic-numbers







share|cite|improve this question






















  • See this Wikipedia article, then consider $ln x= ln 2cdot log_2 x$
    – Arthur
    Nov 23 '18 at 7:59












  • Thanks, Arthur but I'm not clear if you are saying it is true or false. $ln 2 < 0.694$ so that $ln 2 < log_2 x$. The article shows that $ln x < 1 + 1/2 + dots + 1/x$. I am not clear what that says about my question.
    – Larry Freeman
    Nov 23 '18 at 8:07










  • It also shows that $1+ln xgeq1+1/2+cdots$.
    – Arthur
    Nov 23 '18 at 8:10












  • ok. Now I get it. You are saying it is true because for $x > 11, log_2 x > (ln x)frac{10}{7} > ln x + 1$ Thanks.
    – Larry Freeman
    Nov 23 '18 at 8:17










  • And then you can check by hand which cases $xleq 11$ it is true for as well. Exactly.
    – Arthur
    Nov 23 '18 at 8:23


















2














If $x ge 5$, is $1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x$



I believe the answer is yes.



Here is my thinking:



(1) $log_2{5} > 2.32 > 2.284 > 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5}$



(2) Assume up to $x$ that $log_2 x > sumlimits_{i le x}frac{1}{i}$



(3) $(x+1)^{x+1} > {{x+1}choose{x+1}}x^{x+1} + {{x+1}choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$



(4) $(x+1)log_2left(frac{x+1}{x}right) > 1$



(5) $log_2(x+1) - log_2(x) > frac{1}{x+1}$



(6) $log_2(x+1) - log_2(x) + log_2(x) = log_2(x+1) > sumlimits_{i le x+1}frac{1}{i}$






proof-verification inequality logarithms harmonic-numbers







share|cite|improve this question






















  • See this Wikipedia article, then consider $ln x= ln 2cdot log_2 x$
    – Arthur
    Nov 23 '18 at 7:59












  • Thanks, Arthur but I'm not clear if you are saying it is true or false. $ln 2 < 0.694$ so that $ln 2 < log_2 x$. The article shows that $ln x < 1 + 1/2 + dots + 1/x$. I am not clear what that says about my question.
    – Larry Freeman
    Nov 23 '18 at 8:07










  • It also shows that $1+ln xgeq1+1/2+cdots$.
    – Arthur
    Nov 23 '18 at 8:10












  • ok. Now I get it. You are saying it is true because for $x > 11, log_2 x > (ln x)frac{10}{7} > ln x + 1$ Thanks.
    – Larry Freeman
    Nov 23 '18 at 8:17










  • And then you can check by hand which cases $xleq 11$ it is true for as well. Exactly.
    – Arthur
    Nov 23 '18 at 8:23
















2












2








2


1





If $x ge 5$, is $1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x$



I believe the answer is yes.



Here is my thinking:



(1) $log_2{5} > 2.32 > 2.284 > 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5}$



(2) Assume up to $x$ that $log_2 x > sumlimits_{i le x}frac{1}{i}$



(3) $(x+1)^{x+1} > {{x+1}choose{x+1}}x^{x+1} + {{x+1}choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$



(4) $(x+1)log_2left(frac{x+1}{x}right) > 1$



(5) $log_2(x+1) - log_2(x) > frac{1}{x+1}$



(6) $log_2(x+1) - log_2(x) + log_2(x) = log_2(x+1) > sumlimits_{i le x+1}frac{1}{i}$






proof-verification inequality logarithms harmonic-numbers







share|cite|improve this question













If $x ge 5$, is $1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x$



I believe the answer is yes.



Here is my thinking:



(1) $log_2{5} > 2.32 > 2.284 > 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5}$



(2) Assume up to $x$ that $log_2 x > sumlimits_{i le x}frac{1}{i}$



(3) $(x+1)^{x+1} > {{x+1}choose{x+1}}x^{x+1} + {{x+1}choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$



(4) $(x+1)log_2left(frac{x+1}{x}right) > 1$



(5) $log_2(x+1) - log_2(x) > frac{1}{x+1}$



(6) $log_2(x+1) - log_2(x) + log_2(x) = log_2(x+1) > sumlimits_{i le x+1}frac{1}{i}$






proof-verification inequality logarithms harmonic-numbers




proof-verification inequality logarithms harmonic-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 23 '18 at 7:55









Larry FreemanLarry Freeman

3,24121239




3,24121239












  • See this Wikipedia article, then consider $ln x= ln 2cdot log_2 x$
    – Arthur
    Nov 23 '18 at 7:59












  • Thanks, Arthur but I'm not clear if you are saying it is true or false. $ln 2 < 0.694$ so that $ln 2 < log_2 x$. The article shows that $ln x < 1 + 1/2 + dots + 1/x$. I am not clear what that says about my question.
    – Larry Freeman
    Nov 23 '18 at 8:07










  • It also shows that $1+ln xgeq1+1/2+cdots$.
    – Arthur
    Nov 23 '18 at 8:10












  • ok. Now I get it. You are saying it is true because for $x > 11, log_2 x > (ln x)frac{10}{7} > ln x + 1$ Thanks.
    – Larry Freeman
    Nov 23 '18 at 8:17










  • And then you can check by hand which cases $xleq 11$ it is true for as well. Exactly.
    – Arthur
    Nov 23 '18 at 8:23




















  • See this Wikipedia article, then consider $ln x= ln 2cdot log_2 x$
    – Arthur
    Nov 23 '18 at 7:59












  • Thanks, Arthur but I'm not clear if you are saying it is true or false. $ln 2 < 0.694$ so that $ln 2 < log_2 x$. The article shows that $ln x < 1 + 1/2 + dots + 1/x$. I am not clear what that says about my question.
    – Larry Freeman
    Nov 23 '18 at 8:07










  • It also shows that $1+ln xgeq1+1/2+cdots$.
    – Arthur
    Nov 23 '18 at 8:10












  • ok. Now I get it. You are saying it is true because for $x > 11, log_2 x > (ln x)frac{10}{7} > ln x + 1$ Thanks.
    – Larry Freeman
    Nov 23 '18 at 8:17










  • And then you can check by hand which cases $xleq 11$ it is true for as well. Exactly.
    – Arthur
    Nov 23 '18 at 8:23


















See this Wikipedia article, then consider $ln x= ln 2cdot log_2 x$
– Arthur
Nov 23 '18 at 7:59






See this Wikipedia article, then consider $ln x= ln 2cdot log_2 x$
– Arthur
Nov 23 '18 at 7:59














Thanks, Arthur but I'm not clear if you are saying it is true or false. $ln 2 < 0.694$ so that $ln 2 < log_2 x$. The article shows that $ln x < 1 + 1/2 + dots + 1/x$. I am not clear what that says about my question.
– Larry Freeman
Nov 23 '18 at 8:07




Thanks, Arthur but I'm not clear if you are saying it is true or false. $ln 2 < 0.694$ so that $ln 2 < log_2 x$. The article shows that $ln x < 1 + 1/2 + dots + 1/x$. I am not clear what that says about my question.
– Larry Freeman
Nov 23 '18 at 8:07












It also shows that $1+ln xgeq1+1/2+cdots$.
– Arthur
Nov 23 '18 at 8:10






It also shows that $1+ln xgeq1+1/2+cdots$.
– Arthur
Nov 23 '18 at 8:10














ok. Now I get it. You are saying it is true because for $x > 11, log_2 x > (ln x)frac{10}{7} > ln x + 1$ Thanks.
– Larry Freeman
Nov 23 '18 at 8:17




ok. Now I get it. You are saying it is true because for $x > 11, log_2 x > (ln x)frac{10}{7} > ln x + 1$ Thanks.
– Larry Freeman
Nov 23 '18 at 8:17












And then you can check by hand which cases $xleq 11$ it is true for as well. Exactly.
– Arthur
Nov 23 '18 at 8:23






And then you can check by hand which cases $xleq 11$ it is true for as well. Exactly.
– Arthur
Nov 23 '18 at 8:23












3 Answers
3






active

oldest

votes


















2














We have that



$$1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x iff 1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < frac{ln x}{ln 2}$$



$$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right)< ln x$$



then recall that by harmonic series



$$sum_{k=1}^x frac1k simln x+gamma+frac1{2x}$$



then



$$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right) sim ln 2ln x+gammaln 2+frac{ln 2}{2x}stackrel{?}<ln x$$



$$(1-ln 2)ln x> gammaln 2+frac{ln 2}{2x}$$



which is true for $x$ sufficiently large.






share|cite|improve this answer























  • After "harmonic series" your summation has an error (the summand should be $1/k$)
    – YiFan
    Nov 23 '18 at 9:29










  • @YiFan Opssss...yes of course I fix the typo. Thanks!
    – gimusi
    Nov 23 '18 at 9:45



















1














We can use induction as well. If we assume that it's true for $n$:
$$1+frac{1}{2}+...+frac{1}{n}+frac{1}{n+1}<log_2(n)+frac{1}{n+1}$$
So we need to prove that
$$log_2(n)+frac{1}{n+1} < log_2(n+1)$$
$$frac{1}{n+1}< log_2left(1+frac{1}{n}right)$$
$$2^{frac{1}{n+1}}<1+frac{1}{n}$$
$$2<left(1+frac{1}{n}right)^{n+1}$$
$$2<left(1+frac{1}{n}right) left(1+frac{1}{n}right)^n$$
And it's true, because
$$1+frac{1}{n}>1$$
And by Bernoulli's inequality:
$$left(1+frac{1}{n}right)^n geq 1+n frac{1}{n}=2$$
So we just need to find a good starting $n$. Let's check it for $n=6$:
$$1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}<log_2(6)$$
$$frac{9}{20}<log_2({6/4})$$
$$2^{frac{9}{20}}<frac{6}{4}$$
And by Bernoulli's inequality,
$$2^{frac{9}{20}}<1+frac{9}{20}<1+frac{10}{20}=frac{6}{4}$$
So it's true $forall ngeq 6 land n in mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).






share|cite|improve this answer































    1














    Yes because $H_n-log n$ is bounded by $1$ and $log_2n$ is a multiple of $log n$.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      We have that



      $$1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x iff 1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < frac{ln x}{ln 2}$$



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right)< ln x$$



      then recall that by harmonic series



      $$sum_{k=1}^x frac1k simln x+gamma+frac1{2x}$$



      then



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right) sim ln 2ln x+gammaln 2+frac{ln 2}{2x}stackrel{?}<ln x$$



      $$(1-ln 2)ln x> gammaln 2+frac{ln 2}{2x}$$



      which is true for $x$ sufficiently large.






      share|cite|improve this answer























      • After "harmonic series" your summation has an error (the summand should be $1/k$)
        – YiFan
        Nov 23 '18 at 9:29










      • @YiFan Opssss...yes of course I fix the typo. Thanks!
        – gimusi
        Nov 23 '18 at 9:45
















      2














      We have that



      $$1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x iff 1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < frac{ln x}{ln 2}$$



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right)< ln x$$



      then recall that by harmonic series



      $$sum_{k=1}^x frac1k simln x+gamma+frac1{2x}$$



      then



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right) sim ln 2ln x+gammaln 2+frac{ln 2}{2x}stackrel{?}<ln x$$



      $$(1-ln 2)ln x> gammaln 2+frac{ln 2}{2x}$$



      which is true for $x$ sufficiently large.






      share|cite|improve this answer























      • After "harmonic series" your summation has an error (the summand should be $1/k$)
        – YiFan
        Nov 23 '18 at 9:29










      • @YiFan Opssss...yes of course I fix the typo. Thanks!
        – gimusi
        Nov 23 '18 at 9:45














      2












      2








      2






      We have that



      $$1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x iff 1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < frac{ln x}{ln 2}$$



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right)< ln x$$



      then recall that by harmonic series



      $$sum_{k=1}^x frac1k simln x+gamma+frac1{2x}$$



      then



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right) sim ln 2ln x+gammaln 2+frac{ln 2}{2x}stackrel{?}<ln x$$



      $$(1-ln 2)ln x> gammaln 2+frac{ln 2}{2x}$$



      which is true for $x$ sufficiently large.






      share|cite|improve this answer














      We have that



      $$1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < log_2 x iff 1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} < frac{ln x}{ln 2}$$



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right)< ln x$$



      then recall that by harmonic series



      $$sum_{k=1}^x frac1k simln x+gamma+frac1{2x}$$



      then



      $$ln 2left(1 + frac{1}{2} + frac{1}{3} + dots + frac{1}{x} right) sim ln 2ln x+gammaln 2+frac{ln 2}{2x}stackrel{?}<ln x$$



      $$(1-ln 2)ln x> gammaln 2+frac{ln 2}{2x}$$



      which is true for $x$ sufficiently large.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 23 '18 at 9:46

























      answered Nov 23 '18 at 8:42









      gimusigimusi

      1




      1












      • After "harmonic series" your summation has an error (the summand should be $1/k$)
        – YiFan
        Nov 23 '18 at 9:29










      • @YiFan Opssss...yes of course I fix the typo. Thanks!
        – gimusi
        Nov 23 '18 at 9:45


















      • After "harmonic series" your summation has an error (the summand should be $1/k$)
        – YiFan
        Nov 23 '18 at 9:29










      • @YiFan Opssss...yes of course I fix the typo. Thanks!
        – gimusi
        Nov 23 '18 at 9:45
















      After "harmonic series" your summation has an error (the summand should be $1/k$)
      – YiFan
      Nov 23 '18 at 9:29




      After "harmonic series" your summation has an error (the summand should be $1/k$)
      – YiFan
      Nov 23 '18 at 9:29












      @YiFan Opssss...yes of course I fix the typo. Thanks!
      – gimusi
      Nov 23 '18 at 9:45




      @YiFan Opssss...yes of course I fix the typo. Thanks!
      – gimusi
      Nov 23 '18 at 9:45











      1














      We can use induction as well. If we assume that it's true for $n$:
      $$1+frac{1}{2}+...+frac{1}{n}+frac{1}{n+1}<log_2(n)+frac{1}{n+1}$$
      So we need to prove that
      $$log_2(n)+frac{1}{n+1} < log_2(n+1)$$
      $$frac{1}{n+1}< log_2left(1+frac{1}{n}right)$$
      $$2^{frac{1}{n+1}}<1+frac{1}{n}$$
      $$2<left(1+frac{1}{n}right)^{n+1}$$
      $$2<left(1+frac{1}{n}right) left(1+frac{1}{n}right)^n$$
      And it's true, because
      $$1+frac{1}{n}>1$$
      And by Bernoulli's inequality:
      $$left(1+frac{1}{n}right)^n geq 1+n frac{1}{n}=2$$
      So we just need to find a good starting $n$. Let's check it for $n=6$:
      $$1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}<log_2(6)$$
      $$frac{9}{20}<log_2({6/4})$$
      $$2^{frac{9}{20}}<frac{6}{4}$$
      And by Bernoulli's inequality,
      $$2^{frac{9}{20}}<1+frac{9}{20}<1+frac{10}{20}=frac{6}{4}$$
      So it's true $forall ngeq 6 land n in mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).






      share|cite|improve this answer




























        1














        We can use induction as well. If we assume that it's true for $n$:
        $$1+frac{1}{2}+...+frac{1}{n}+frac{1}{n+1}<log_2(n)+frac{1}{n+1}$$
        So we need to prove that
        $$log_2(n)+frac{1}{n+1} < log_2(n+1)$$
        $$frac{1}{n+1}< log_2left(1+frac{1}{n}right)$$
        $$2^{frac{1}{n+1}}<1+frac{1}{n}$$
        $$2<left(1+frac{1}{n}right)^{n+1}$$
        $$2<left(1+frac{1}{n}right) left(1+frac{1}{n}right)^n$$
        And it's true, because
        $$1+frac{1}{n}>1$$
        And by Bernoulli's inequality:
        $$left(1+frac{1}{n}right)^n geq 1+n frac{1}{n}=2$$
        So we just need to find a good starting $n$. Let's check it for $n=6$:
        $$1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}<log_2(6)$$
        $$frac{9}{20}<log_2({6/4})$$
        $$2^{frac{9}{20}}<frac{6}{4}$$
        And by Bernoulli's inequality,
        $$2^{frac{9}{20}}<1+frac{9}{20}<1+frac{10}{20}=frac{6}{4}$$
        So it's true $forall ngeq 6 land n in mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).






        share|cite|improve this answer


























          1












          1








          1






          We can use induction as well. If we assume that it's true for $n$:
          $$1+frac{1}{2}+...+frac{1}{n}+frac{1}{n+1}<log_2(n)+frac{1}{n+1}$$
          So we need to prove that
          $$log_2(n)+frac{1}{n+1} < log_2(n+1)$$
          $$frac{1}{n+1}< log_2left(1+frac{1}{n}right)$$
          $$2^{frac{1}{n+1}}<1+frac{1}{n}$$
          $$2<left(1+frac{1}{n}right)^{n+1}$$
          $$2<left(1+frac{1}{n}right) left(1+frac{1}{n}right)^n$$
          And it's true, because
          $$1+frac{1}{n}>1$$
          And by Bernoulli's inequality:
          $$left(1+frac{1}{n}right)^n geq 1+n frac{1}{n}=2$$
          So we just need to find a good starting $n$. Let's check it for $n=6$:
          $$1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}<log_2(6)$$
          $$frac{9}{20}<log_2({6/4})$$
          $$2^{frac{9}{20}}<frac{6}{4}$$
          And by Bernoulli's inequality,
          $$2^{frac{9}{20}}<1+frac{9}{20}<1+frac{10}{20}=frac{6}{4}$$
          So it's true $forall ngeq 6 land n in mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).






          share|cite|improve this answer














          We can use induction as well. If we assume that it's true for $n$:
          $$1+frac{1}{2}+...+frac{1}{n}+frac{1}{n+1}<log_2(n)+frac{1}{n+1}$$
          So we need to prove that
          $$log_2(n)+frac{1}{n+1} < log_2(n+1)$$
          $$frac{1}{n+1}< log_2left(1+frac{1}{n}right)$$
          $$2^{frac{1}{n+1}}<1+frac{1}{n}$$
          $$2<left(1+frac{1}{n}right)^{n+1}$$
          $$2<left(1+frac{1}{n}right) left(1+frac{1}{n}right)^n$$
          And it's true, because
          $$1+frac{1}{n}>1$$
          And by Bernoulli's inequality:
          $$left(1+frac{1}{n}right)^n geq 1+n frac{1}{n}=2$$
          So we just need to find a good starting $n$. Let's check it for $n=6$:
          $$1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}<log_2(6)$$
          $$frac{9}{20}<log_2({6/4})$$
          $$2^{frac{9}{20}}<frac{6}{4}$$
          And by Bernoulli's inequality,
          $$2^{frac{9}{20}}<1+frac{9}{20}<1+frac{10}{20}=frac{6}{4}$$
          So it's true $forall ngeq 6 land n in mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).







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          edited Nov 23 '18 at 10:51

























          answered Nov 23 '18 at 10:34









          BotondBotond

          5,5882732




          5,5882732























              1














              Yes because $H_n-log n$ is bounded by $1$ and $log_2n$ is a multiple of $log n$.






              share|cite|improve this answer


























                1














                Yes because $H_n-log n$ is bounded by $1$ and $log_2n$ is a multiple of $log n$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  Yes because $H_n-log n$ is bounded by $1$ and $log_2n$ is a multiple of $log n$.






                  share|cite|improve this answer












                  Yes because $H_n-log n$ is bounded by $1$ and $log_2n$ is a multiple of $log n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 '18 at 10:58









                  Yves DaoustYves Daoust

                  124k671222




                  124k671222






























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