Inequality 1 n + 1 + 1 n + 2 + . . . + 1 3 n + 1 &gt; 1This is a 9th grade problem. I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only 2 n + 1 terms. Right? After that I have no idea. Thx!

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Inequality        1          n      +      1        +      1          n      +      2        +  .  .  .  +      1          3      n      +      1        &amp;gt;  1This is a 9th grade problem. I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only   2  n  +  1 terms. Right? After that I have no idea. Thx!

Miah Carlson · Accepted Answer

There are   2  n  +  1 terms in the sum, you just need to pair up the terms symmetrically from both ends, take average and compare with the term in the middle.                              ∑                      k            =            n            +            1                                3            n            +            1                                    1          k                                    =                  ∑                      k            =            −            n                    n                          1                      2            n            +            1            +            k                          =                  1          2                          ∑                      k            =            −            n                    n                          (                      1                          2              n              +              1              +              k                                +                      1                          2              n              +              1              −              k                                )                                                  =                  ∑                      k            =            −            n                    n                                      2            n            +            1                                (            2            n            +            1                          )              2                        −                          k              2                                                                          &amp;gt;                                                       assume                 n                &amp;gt;                0                                                              ∑                      k            =            −            n                    n                          1                      2            n            +            1                          =                              2            n            +            1                                2            n            +            1                          =        1

Zackary Diaz · Accepted Answer

By C-S      ∑          k      =      1              2      n      +      1            1          n      +      k        ≥            (      2      n      +      1              )        2                            ∑                  k          =          1                          2          n          +          1                    (      n      +      k      )        =            (      2      n      +      1              )        2                            (        2        (        n        +        1        )        +        2        n        )        (        2        n        +        1        )            2        =  1.

Inequality 1/(n+1)+1/(n+2)+...+1/(3n+1)>1,AA n in NN This is a 9th grade problem. I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only 2n+1 terms. Right? After that I have no idea.

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