Prove that $prodlimits_{k=1}^infty left(1+frac1{2^k}right) lt e ?$












1














How would you prove that $$displaystyle prod_{k=1}^infty left(1+dfrac{1}{2^k}right) lt e ?$$



Wolfram|Alpha shows that the product evaluates to $2.384231 dots$ but is there a nice way to write this number?



A hint about solving the problem was given but I don't know how to prove the lemma.



Lemma : Let, $a_1,a_2,a_3, ldots,a_n$ be positive numbers and let $s=a_1+a_2+a_3+cdots+a_n$ then $$(1+a_1)(1+a_2)(1+a_3)cdots(1+a_n)$$ $$le 1+s+dfrac{s^2}{2!}+dfrac{s^3}{3!}+cdots+dfrac{s^n}{n!}$$






inequality products infinite-product







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  • Can you figure out a way to complete the proof by the given hint?
    – gimusi
    Oct 22 '18 at 9:46










  • math.stackexchange.com/questions/1110242/…
    – Jack D'Aurizio
    Oct 22 '18 at 13:12
















1














How would you prove that $$displaystyle prod_{k=1}^infty left(1+dfrac{1}{2^k}right) lt e ?$$



Wolfram|Alpha shows that the product evaluates to $2.384231 dots$ but is there a nice way to write this number?



A hint about solving the problem was given but I don't know how to prove the lemma.



Lemma : Let, $a_1,a_2,a_3, ldots,a_n$ be positive numbers and let $s=a_1+a_2+a_3+cdots+a_n$ then $$(1+a_1)(1+a_2)(1+a_3)cdots(1+a_n)$$ $$le 1+s+dfrac{s^2}{2!}+dfrac{s^3}{3!}+cdots+dfrac{s^n}{n!}$$






inequality products infinite-product







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  • Can you figure out a way to complete the proof by the given hint?
    – gimusi
    Oct 22 '18 at 9:46










  • math.stackexchange.com/questions/1110242/…
    – Jack D'Aurizio
    Oct 22 '18 at 13:12














1












1








1


1





How would you prove that $$displaystyle prod_{k=1}^infty left(1+dfrac{1}{2^k}right) lt e ?$$



Wolfram|Alpha shows that the product evaluates to $2.384231 dots$ but is there a nice way to write this number?



A hint about solving the problem was given but I don't know how to prove the lemma.



Lemma : Let, $a_1,a_2,a_3, ldots,a_n$ be positive numbers and let $s=a_1+a_2+a_3+cdots+a_n$ then $$(1+a_1)(1+a_2)(1+a_3)cdots(1+a_n)$$ $$le 1+s+dfrac{s^2}{2!}+dfrac{s^3}{3!}+cdots+dfrac{s^n}{n!}$$






inequality products infinite-product







share|cite|improve this question















How would you prove that $$displaystyle prod_{k=1}^infty left(1+dfrac{1}{2^k}right) lt e ?$$



Wolfram|Alpha shows that the product evaluates to $2.384231 dots$ but is there a nice way to write this number?



A hint about solving the problem was given but I don't know how to prove the lemma.



Lemma : Let, $a_1,a_2,a_3, ldots,a_n$ be positive numbers and let $s=a_1+a_2+a_3+cdots+a_n$ then $$(1+a_1)(1+a_2)(1+a_3)cdots(1+a_n)$$ $$le 1+s+dfrac{s^2}{2!}+dfrac{s^3}{3!}+cdots+dfrac{s^n}{n!}$$






inequality products infinite-product




inequality products infinite-product






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edited Dec 1 '18 at 21:10









Martin Sleziak

44.7k7115270




44.7k7115270










asked Oct 22 '18 at 8:18









AtiqRahman

191




191












  • Can you figure out a way to complete the proof by the given hint?
    – gimusi
    Oct 22 '18 at 9:46










  • math.stackexchange.com/questions/1110242/…
    – Jack D'Aurizio
    Oct 22 '18 at 13:12


















  • Can you figure out a way to complete the proof by the given hint?
    – gimusi
    Oct 22 '18 at 9:46










  • math.stackexchange.com/questions/1110242/…
    – Jack D'Aurizio
    Oct 22 '18 at 13:12
















Can you figure out a way to complete the proof by the given hint?
– gimusi
Oct 22 '18 at 9:46




Can you figure out a way to complete the proof by the given hint?
– gimusi
Oct 22 '18 at 9:46












math.stackexchange.com/questions/1110242/…
– Jack D'Aurizio
Oct 22 '18 at 13:12




math.stackexchange.com/questions/1110242/…
– Jack D'Aurizio
Oct 22 '18 at 13:12










3 Answers
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3














HINT



Taking $log$ both sides the statement is equivalent to prove that



$$sum_{k=1}^infty log left(1+dfrac{1}{2^k}right) lt 1$$



then use $log(1+x)<x$.






share|cite|improve this answer





























    2














    It can be easily shown that for any $xin(0,1)$ we have



    $$ frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9765}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9765}} <1+x < frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9450}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9450}} $$
    and in general, by setting $D_m=prod_{k=1}^{m}(2^k-1)$ and $p_n(x)=1+sum_{k=1}^{n}frac{x^k}{D_k}$,
    $$ frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+frac{x^{n+1}}{D_{n+1}}}<1+x< frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$
    By telescoping it follows that
    $$ p_n(1)+frac{1}{D_{n+1}}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<p_n(1)+frac{1}{D_n(2^{n+1}-2)} $$
    and by picking $n=4$ we have
    $$ frac{3326}{1395}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<frac{22531}{9450} $$
    such that the first figures of the middle term are $color{green}{2.3842}$.
    By picking $n=10$ we get that the middle term is $color{green}{2.384231029ldots}$.

    As a continued fraction
    $$ prod_{kgeq 1}left(1+frac{1}{2^k}right)=left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,ldotsright]$$
    while $e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots]$.






    share|cite|improve this answer





























      0














      Proof of the Lemma by Induction.
      Let $P(n): (1+a_1)(1+a_2)...(1+a_n) ≤ 1 + s_n + ... + frac {s_n^n}{n!}$
      For$n=1$ the result is obvious.
      Let $P(m): (1+a_1)(1+a_2)...(1+a_m) ≤ 1 + s_m + ... + frac {s_m^m}{m!}$ be true. Then,
      $$1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}$$
      $$= 1 + (s_m + a_{m+1}) + ... + frac {(s_m + a_{m+1})^{m+1}}{(m+1)!}$$
      $$= 1 + s_m + ... + frac {s_m^m}{m!} + Q , Q>0$$
      Thus we have,
      $$(1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}) - Q = 1 + s_m + ... + frac {s_m^m}{m!} ≥ (1+a_1)(1+a_2)...(1+a_m) < (1+a_1)(1+a_2)...(1+a_m)(1+a_{m+1})$$
      Hence $P(m+1)$ is true whenever $P(m)$ is true.
      Therefore $P(n)$ is true in general.






      share|cite|improve this answer























      • Induction doesn't give much insight for the proof.
        – AtiqRahman
        Oct 22 '18 at 8:49











      Your Answer





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      3 Answers
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      3 Answers
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      3














      HINT



      Taking $log$ both sides the statement is equivalent to prove that



      $$sum_{k=1}^infty log left(1+dfrac{1}{2^k}right) lt 1$$



      then use $log(1+x)<x$.






      share|cite|improve this answer


























        3














        HINT



        Taking $log$ both sides the statement is equivalent to prove that



        $$sum_{k=1}^infty log left(1+dfrac{1}{2^k}right) lt 1$$



        then use $log(1+x)<x$.






        share|cite|improve this answer
























          3












          3








          3






          HINT



          Taking $log$ both sides the statement is equivalent to prove that



          $$sum_{k=1}^infty log left(1+dfrac{1}{2^k}right) lt 1$$



          then use $log(1+x)<x$.






          share|cite|improve this answer












          HINT



          Taking $log$ both sides the statement is equivalent to prove that



          $$sum_{k=1}^infty log left(1+dfrac{1}{2^k}right) lt 1$$



          then use $log(1+x)<x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 22 '18 at 8:27









          gimusi

          1




          1























              2














              It can be easily shown that for any $xin(0,1)$ we have



              $$ frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9765}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9765}} <1+x < frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9450}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9450}} $$
              and in general, by setting $D_m=prod_{k=1}^{m}(2^k-1)$ and $p_n(x)=1+sum_{k=1}^{n}frac{x^k}{D_k}$,
              $$ frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+frac{x^{n+1}}{D_{n+1}}}<1+x< frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$
              By telescoping it follows that
              $$ p_n(1)+frac{1}{D_{n+1}}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<p_n(1)+frac{1}{D_n(2^{n+1}-2)} $$
              and by picking $n=4$ we have
              $$ frac{3326}{1395}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<frac{22531}{9450} $$
              such that the first figures of the middle term are $color{green}{2.3842}$.
              By picking $n=10$ we get that the middle term is $color{green}{2.384231029ldots}$.

              As a continued fraction
              $$ prod_{kgeq 1}left(1+frac{1}{2^k}right)=left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,ldotsright]$$
              while $e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots]$.






              share|cite|improve this answer


























                2














                It can be easily shown that for any $xin(0,1)$ we have



                $$ frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9765}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9765}} <1+x < frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9450}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9450}} $$
                and in general, by setting $D_m=prod_{k=1}^{m}(2^k-1)$ and $p_n(x)=1+sum_{k=1}^{n}frac{x^k}{D_k}$,
                $$ frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+frac{x^{n+1}}{D_{n+1}}}<1+x< frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$
                By telescoping it follows that
                $$ p_n(1)+frac{1}{D_{n+1}}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<p_n(1)+frac{1}{D_n(2^{n+1}-2)} $$
                and by picking $n=4$ we have
                $$ frac{3326}{1395}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<frac{22531}{9450} $$
                such that the first figures of the middle term are $color{green}{2.3842}$.
                By picking $n=10$ we get that the middle term is $color{green}{2.384231029ldots}$.

                As a continued fraction
                $$ prod_{kgeq 1}left(1+frac{1}{2^k}right)=left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,ldotsright]$$
                while $e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots]$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  It can be easily shown that for any $xin(0,1)$ we have



                  $$ frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9765}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9765}} <1+x < frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9450}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9450}} $$
                  and in general, by setting $D_m=prod_{k=1}^{m}(2^k-1)$ and $p_n(x)=1+sum_{k=1}^{n}frac{x^k}{D_k}$,
                  $$ frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+frac{x^{n+1}}{D_{n+1}}}<1+x< frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$
                  By telescoping it follows that
                  $$ p_n(1)+frac{1}{D_{n+1}}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<p_n(1)+frac{1}{D_n(2^{n+1}-2)} $$
                  and by picking $n=4$ we have
                  $$ frac{3326}{1395}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<frac{22531}{9450} $$
                  such that the first figures of the middle term are $color{green}{2.3842}$.
                  By picking $n=10$ we get that the middle term is $color{green}{2.384231029ldots}$.

                  As a continued fraction
                  $$ prod_{kgeq 1}left(1+frac{1}{2^k}right)=left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,ldotsright]$$
                  while $e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots]$.






                  share|cite|improve this answer












                  It can be easily shown that for any $xin(0,1)$ we have



                  $$ frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9765}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9765}} <1+x < frac{1+2x+frac{4x^2}{3}+frac{8x^3}{21}+frac{16 x^4}{315}+frac{32 x^5}{9450}}{1+x+frac{x^2}{3}+frac{x^3}{21}+frac{x^4}{315}+frac{x^5}{9450}} $$
                  and in general, by setting $D_m=prod_{k=1}^{m}(2^k-1)$ and $p_n(x)=1+sum_{k=1}^{n}frac{x^k}{D_k}$,
                  $$ frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+frac{x^{n+1}}{D_{n+1}}}<1+x< frac{p_n(2x)+frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$
                  By telescoping it follows that
                  $$ p_n(1)+frac{1}{D_{n+1}}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<p_n(1)+frac{1}{D_n(2^{n+1}-2)} $$
                  and by picking $n=4$ we have
                  $$ frac{3326}{1395}<prod_{kgeq 1}left(1+frac{1}{2^k}right)<frac{22531}{9450} $$
                  such that the first figures of the middle term are $color{green}{2.3842}$.
                  By picking $n=10$ we get that the middle term is $color{green}{2.384231029ldots}$.

                  As a continued fraction
                  $$ prod_{kgeq 1}left(1+frac{1}{2^k}right)=left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,ldotsright]$$
                  while $e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots]$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 22 '18 at 13:51









                  Jack D'Aurizio

                  286k33279656




                  286k33279656























                      0














                      Proof of the Lemma by Induction.
                      Let $P(n): (1+a_1)(1+a_2)...(1+a_n) ≤ 1 + s_n + ... + frac {s_n^n}{n!}$
                      For$n=1$ the result is obvious.
                      Let $P(m): (1+a_1)(1+a_2)...(1+a_m) ≤ 1 + s_m + ... + frac {s_m^m}{m!}$ be true. Then,
                      $$1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}$$
                      $$= 1 + (s_m + a_{m+1}) + ... + frac {(s_m + a_{m+1})^{m+1}}{(m+1)!}$$
                      $$= 1 + s_m + ... + frac {s_m^m}{m!} + Q , Q>0$$
                      Thus we have,
                      $$(1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}) - Q = 1 + s_m + ... + frac {s_m^m}{m!} ≥ (1+a_1)(1+a_2)...(1+a_m) < (1+a_1)(1+a_2)...(1+a_m)(1+a_{m+1})$$
                      Hence $P(m+1)$ is true whenever $P(m)$ is true.
                      Therefore $P(n)$ is true in general.






                      share|cite|improve this answer























                      • Induction doesn't give much insight for the proof.
                        – AtiqRahman
                        Oct 22 '18 at 8:49
















                      0














                      Proof of the Lemma by Induction.
                      Let $P(n): (1+a_1)(1+a_2)...(1+a_n) ≤ 1 + s_n + ... + frac {s_n^n}{n!}$
                      For$n=1$ the result is obvious.
                      Let $P(m): (1+a_1)(1+a_2)...(1+a_m) ≤ 1 + s_m + ... + frac {s_m^m}{m!}$ be true. Then,
                      $$1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}$$
                      $$= 1 + (s_m + a_{m+1}) + ... + frac {(s_m + a_{m+1})^{m+1}}{(m+1)!}$$
                      $$= 1 + s_m + ... + frac {s_m^m}{m!} + Q , Q>0$$
                      Thus we have,
                      $$(1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}) - Q = 1 + s_m + ... + frac {s_m^m}{m!} ≥ (1+a_1)(1+a_2)...(1+a_m) < (1+a_1)(1+a_2)...(1+a_m)(1+a_{m+1})$$
                      Hence $P(m+1)$ is true whenever $P(m)$ is true.
                      Therefore $P(n)$ is true in general.






                      share|cite|improve this answer























                      • Induction doesn't give much insight for the proof.
                        – AtiqRahman
                        Oct 22 '18 at 8:49














                      0












                      0








                      0






                      Proof of the Lemma by Induction.
                      Let $P(n): (1+a_1)(1+a_2)...(1+a_n) ≤ 1 + s_n + ... + frac {s_n^n}{n!}$
                      For$n=1$ the result is obvious.
                      Let $P(m): (1+a_1)(1+a_2)...(1+a_m) ≤ 1 + s_m + ... + frac {s_m^m}{m!}$ be true. Then,
                      $$1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}$$
                      $$= 1 + (s_m + a_{m+1}) + ... + frac {(s_m + a_{m+1})^{m+1}}{(m+1)!}$$
                      $$= 1 + s_m + ... + frac {s_m^m}{m!} + Q , Q>0$$
                      Thus we have,
                      $$(1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}) - Q = 1 + s_m + ... + frac {s_m^m}{m!} ≥ (1+a_1)(1+a_2)...(1+a_m) < (1+a_1)(1+a_2)...(1+a_m)(1+a_{m+1})$$
                      Hence $P(m+1)$ is true whenever $P(m)$ is true.
                      Therefore $P(n)$ is true in general.






                      share|cite|improve this answer














                      Proof of the Lemma by Induction.
                      Let $P(n): (1+a_1)(1+a_2)...(1+a_n) ≤ 1 + s_n + ... + frac {s_n^n}{n!}$
                      For$n=1$ the result is obvious.
                      Let $P(m): (1+a_1)(1+a_2)...(1+a_m) ≤ 1 + s_m + ... + frac {s_m^m}{m!}$ be true. Then,
                      $$1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}$$
                      $$= 1 + (s_m + a_{m+1}) + ... + frac {(s_m + a_{m+1})^{m+1}}{(m+1)!}$$
                      $$= 1 + s_m + ... + frac {s_m^m}{m!} + Q , Q>0$$
                      Thus we have,
                      $$(1 + s_{m+1} + ... + frac {s_{m+1}^{m+1}}{(m+1)!}) - Q = 1 + s_m + ... + frac {s_m^m}{m!} ≥ (1+a_1)(1+a_2)...(1+a_m) < (1+a_1)(1+a_2)...(1+a_m)(1+a_{m+1})$$
                      Hence $P(m+1)$ is true whenever $P(m)$ is true.
                      Therefore $P(n)$ is true in general.







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                      edited Oct 22 '18 at 8:51

























                      answered Oct 22 '18 at 8:47









                      Awe Kumar Jha

                      38613




                      38613












                      • Induction doesn't give much insight for the proof.
                        – AtiqRahman
                        Oct 22 '18 at 8:49


















                      • Induction doesn't give much insight for the proof.
                        – AtiqRahman
                        Oct 22 '18 at 8:49
















                      Induction doesn't give much insight for the proof.
                      – AtiqRahman
                      Oct 22 '18 at 8:49




                      Induction doesn't give much insight for the proof.
                      – AtiqRahman
                      Oct 22 '18 at 8:49


















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