Prove inequality by induction ∀ n ∈ N , n ≥ 2 : ( 1 + 1 n ) n &lt; ∑ k = 0 n 1 k ! &lt; 3I have the following statement to prove: ∀ n ∈ N , n ≥ 2 : ( 1 + 1 n ) n &lt; ∑ k = 0 n 1 k ! &lt; 3What I have already proven is that ( 1 + 1 n ) n &lt; ( 1 + 1 n + 1 ) n + 1 I have started, as usual by induction with n = 2. Then I went on to say ( 1 + 1 n + 1 ) ∗ ( 1 + 1 n + 1 ) n &lt; ∑ k = 0 n 1 k ! + 1 ( n + 1 ) ! &lt; 3But this is where I seem to get stuck. I cannot use the induction hypothesis, since the denominator on the left side is now n + 1 instead of n. I assume, I have to use the other inequality, which I have already proven to be true, but I do not know how. Any hints?

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Prove inequality by induction   ∀  n  ∈      N    ,  n  ≥  2        :        (  1  +      1    n        )    n    &amp;lt;      ∑          k      =      0        n        1          k      !        &amp;lt;  3I have the following statement to prove:  ∀  n  ∈      N    ,  n  ≥  2        :        (  1  +      1    n        )    n    &amp;lt;      ∑          k      =      0        n        1          k      !        &amp;lt;  3What I have already proven is that  (  1  +      1    n        )    n    &amp;lt;  (  1  +      1          n      +      1            )          n      +      1      I have started, as usual by induction with   n  =  2. Then I went on to say  (  1  +      1          n      +      1        )  ∗  (  1  +      1          n      +      1            )          n        &amp;lt;      ∑          k      =      0        n        1          k      !        +      1          (      n      +      1      )      !        &amp;lt;  3But this is where I seem to get stuck. I cannot use the induction hypothesis, since the denominator on the left side is now   n  +  1 instead of   n. I assume, I have to use the other inequality, which I have already proven to be true, but I do not know how. Any hints?

Mateo Barajas · Accepted Answer

By Binom Newton            (      1      +              1        n            )        n    =  2  +            1      −              1        n                    2      !        +                    (        1        −                  1          n                )                    (        1        −                  2          n                )                    3      !        +  .  .  .  +                    (        1        −                  1          n                )                    (        1        −                  2          n                )            .      .      .              (        1        −                              n            −            1                    n                )                    n      !        &amp;lt;  &amp;lt;  2  +      1          2      !        +      1          3      !        .  .  .  +      1          n      !        &amp;lt;  2  +      1    2    +      1          2      2        +  .  .  .  =  3

Prove inequality by induction <mi mathvariant="normal">∀ n ∈<!-- ∈

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