Let $K$ a operator of continued fraction.Is this true?

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$DeclareMathOperator*KK$
Let $K$ a operator of continued fraction.
Show that:
$$1+sum_k=1^n(-1)^kvarphi_4cdots varphi_2k+2=frac11+displaystyleK_k=2^n+1fracvarphi_2k1-varphi_2k$$



What's a name of this little theorem? This theorem is true?







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  • In other words, $1+sum_k=1^n(-1)^kprod_j=2^k+1a_j=frac11+displaystyleK_k=2^n+1fraca_k1-a_k $.
    – marty cohen
    Jul 16 at 19:48










  • Probably Euler.
    – marty cohen
    Jul 16 at 19:49














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$DeclareMathOperator*KK$
Let $K$ a operator of continued fraction.
Show that:
$$1+sum_k=1^n(-1)^kvarphi_4cdots varphi_2k+2=frac11+displaystyleK_k=2^n+1fracvarphi_2k1-varphi_2k$$



What's a name of this little theorem? This theorem is true?







share|cite|improve this question





















  • In other words, $1+sum_k=1^n(-1)^kprod_j=2^k+1a_j=frac11+displaystyleK_k=2^n+1fraca_k1-a_k $.
    – marty cohen
    Jul 16 at 19:48










  • Probably Euler.
    – marty cohen
    Jul 16 at 19:49












up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





$DeclareMathOperator*KK$
Let $K$ a operator of continued fraction.
Show that:
$$1+sum_k=1^n(-1)^kvarphi_4cdots varphi_2k+2=frac11+displaystyleK_k=2^n+1fracvarphi_2k1-varphi_2k$$



What's a name of this little theorem? This theorem is true?







share|cite|improve this question













$DeclareMathOperator*KK$
Let $K$ a operator of continued fraction.
Show that:
$$1+sum_k=1^n(-1)^kvarphi_4cdots varphi_2k+2=frac11+displaystyleK_k=2^n+1fracvarphi_2k1-varphi_2k$$



What's a name of this little theorem? This theorem is true?









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 19:41









Fabio Lucchini

5,62411025




5,62411025









asked Jul 16 at 17:38









Israel Meireles Chrisostomo

562313




562313











  • In other words, $1+sum_k=1^n(-1)^kprod_j=2^k+1a_j=frac11+displaystyleK_k=2^n+1fraca_k1-a_k $.
    – marty cohen
    Jul 16 at 19:48










  • Probably Euler.
    – marty cohen
    Jul 16 at 19:49
















  • In other words, $1+sum_k=1^n(-1)^kprod_j=2^k+1a_j=frac11+displaystyleK_k=2^n+1fraca_k1-a_k $.
    – marty cohen
    Jul 16 at 19:48










  • Probably Euler.
    – marty cohen
    Jul 16 at 19:49















In other words, $1+sum_k=1^n(-1)^kprod_j=2^k+1a_j=frac11+displaystyleK_k=2^n+1fraca_k1-a_k $.
– marty cohen
Jul 16 at 19:48




In other words, $1+sum_k=1^n(-1)^kprod_j=2^k+1a_j=frac11+displaystyleK_k=2^n+1fraca_k1-a_k $.
– marty cohen
Jul 16 at 19:48












Probably Euler.
– marty cohen
Jul 16 at 19:49




Probably Euler.
– marty cohen
Jul 16 at 19:49















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