Problem proving inequality 4 n n + 1 ≤ ( 2 n ) ! ( n ! ) 2 I skip the base case n = 0 because it's obvious.I know that this is very equivalent to: Prove by induction: 2 2 n n + 1 &lt; ( 2 n ) ! ( n ! ) 2 , n &gt; 1But I try to learn some tricks so maybe you can help me?Assumption: 4 n n + 1 ≤ ( 2 n ) ! ( n ! ) 2 ⟺ 4 n ≤ ( n + 1 ) ( 2 n ) ! ( n ! ) 2 for some n.Step: n → n + 1 4 n + 1 = ( n + 2 ) ( 2 n + 2 ) ! ( ( n + 1 ) ! ) 2 I begin with manipulating the LHS: 4 n + 1 = 4 ⋅ 4 n ≤ ? ? ? ? ? ? ≤ ( 2 n ) ! ( 2 n + 2 ) ( 2 n + 1 ) ( n + 1 ) ( n ! ( n + 1 ) ) = ( n + 1 ) ( 2 n + 2 ) ! ( ( n + 1 ) ! ) 2 Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.Can you help me?

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Problem proving inequality             4      n              n      +      1        ≤            (      2      n      )      !              (      n      !              )        2            I skip the base case   n  =  0 because it&#039;s obvious.I know that this is very equivalent to: Prove by induction:             2              2        n                    n      +      1        &amp;lt;            (      2      n      )      !              (      n      !              )        2              ,  n  &amp;gt;  1But I try to learn some tricks so maybe you can help me?Assumption:             4      n              n      +      1        ≤            (      2      n      )      !              (      n      !              )        2                ⟺        4    n    ≤            (      n      +      1      )      (      2      n      )      !              (      n      !              )        2             for some n.Step:   n  →  n  +  1      4          n      +      1        =            (      n      +      2      )      (      2      n      +      2      )      !              (      (      n      +      1      )      !              )        2            I begin with manipulating the LHS:      4          n      +      1        =  4  ⋅      4    n    ≤  ?  ?  ?  ?  ?  ?  ≤            (      2      n      )      !      (      2      n      +      2      )      (      2      n      +      1      )      (      n      +      1      )              (      n      !      (      n      +      1      )      )        =            (      n      +      1      )      (      2      n      +      2      )      !              (      (      n      +      1      )      !              )        2            Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.Can you help me?

scranna0o · Accepted Answer

4              n        +        1                    (      n      +      2      )        &amp;lt;            (      2      n      )      !              (      n      !              )        2                        4      (      n      +      1      )              (      n      +      2      )      Now            (      2      n      )      !              (      n      !              )        2                        4      (      n      +      1      )              (      n      +      2      )        =            (      2      n      +      2      )      !              (      (      n      +      1      )      !              )        2                        4      (      n      +      1              )        3                    (      n      +      2      )      (      2      n      +      1      )      (      2      n      +      2      )      So            4      (      n      +      1              )        3                    (      n      +      2      )      (      2      n      +      1      )      (      2      n      +      2      )        &amp;lt;  1implies            (      2      n      +      2      )      !              (      (      n      +      1      )      !              )        2                        4      (      n      +      1              )        3                    (      n      +      2      )      (      2      n      +      1      )      (      2      n      +      2      )        &amp;lt;            (      2      n      +      2      )      !              (      (      n      +      1      )      !              )        2            , for   n  &amp;gt;  1

Problem proving inequality (4^n)/(n+1)<=((2n)!)/((n!)^2) I skip the base case n=0 because it's obvious. I know that this is very equivalent to: Prove by induction: (2^(2n))/(n+1)<((2n)!)/((n!)^2),n>1

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