Want to understand how a fraction is simplifiedThe fraction is used to determine the sum of a telescopic series. ∑ k = 1 ∞ 1 ( k + 1 ) k + k k + 1 This is the solved fraction. 1 ( k + 1 ) k + k k + 1 = ( k + 1 ) k − k k + 1 ( k + 1 ) 2 k − k 2 ( k + 1 ) = ( k + 1 ) k − k k + 1 k 3 + 2 k 2 + k − k 3 − k 2 = ( k + 1 ) k − k k + 1 k ( k + 1 ) = k k − k + 1 k + 1 = 1 k − 1 k + 1 I want to understand what is being done on every step,mainly the last three. Thank you.

Question

Want to understand how a fraction is simplifiedThe fraction is used to determine the sum of a telescopic series.      ∑          k      =      1        ∞        1          (      k      +      1      )              k            +      k              k        +        1            This is the solved fraction.      1          (      k      +      1      )              k            +      k              k        +        1              =            (      k      +      1      )              k            −      k              k        +        1                    (      k      +      1              )        2            k      −              k        2            (      k      +      1      )        =            (      k      +      1      )              k            −      k              k        +        1                            k        3            +      2              k        2            +      k      −              k        3            −              k        2              =            (      k      +      1      )              k            −      k              k        +        1                    k      (      k      +      1      )        =            k        k    −            k      +      1              k      +      1        =      1          k        −      1          k      +      1      I want to understand what is being done on every step,mainly the last three. Thank you.

Ashlee Ramos · Accepted Answer

First equal: multiply and divide by   (  k  +  1  )      k    −  k      k    +    1  Second equal: Expand the denominatorThird equal: cancel obvious terms in the denominator and write       k    2    +  k  =  k  (  k  +  1  ).Fourth equal: distribute along the minus in the numerator, and cancel the obvious factors   (  k  +  1  ) in the first summand, and k in the second one:            (      k      +      1      )              k            −      k              k        +        1                    k      (      k      +      1      )        =            (      k      +      1      )              k                    k      (      k      +      1      )        −            k              k        +        1                    k      (      k      +      1      )        =            k        k    −            k      +      1              k      +      1        .Fifth equal:             k        k    =      1          k      , and similarly for   k  +  1

tuzkutimonq4 · Accepted Answer

Another way to look at it is first &quot;declutter&quot; the expression, since all those radicals obfuscate the simple structure. Let   a  =      k    ,   b  =      k    +    1    , then:      1          (      k      +      1      )              k            +      k              k        +        1              =      1          a              b        2            +              a        2            b        =      1          a      b      (      a      +      b      )      Now consider that     (  b  +  a  )  (  b  −  a  )  =      b    2    −      a    2    =  (      k    +  1  )  −      k    =  1  , so               1              a        +        b              =  b  −  a  . Then:      1          a      b      (      a      +      b      )        =                    b        −        a                    a        b              =                    b                    a                  b                      −                    a                              a                b              =            1      a        −            1      b        =            1              k              −            1              k        +        1

Want to understand how a fraction is simplified The fraction is used to determine the sum of a telescopic series. sum_(k=1)^oo 1/((k+1)sqrt(k)+k sqrt(k+1))

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