Question 1133418
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The requested sum is  (it is how I read the problem)


{{{2/(1*2)}}} + {{{2/(2*3)}}} + {{{2/(3*4)}}} + . . . {{{2/(n*(n+1))}}} + . . . to infinity = 


        Each term  {{{2/(n*(n+1))}}} = {{{2/n}}} - {{{2/(n+1)}}}.



        Therefore, any finite sum   {{{2/(1*2)}}} + {{{2/(2*3)}}} + {{{2/(3*4)}}} + . . . {{{2/(n*(n+1))}}} = 


        = ({{{2/1}}} - {{{2/2}}}) + ({{{2/2}}} - {{{2/3}}}) + ({{{2/3}}} - {{{3/4}}}) + . . . + ({{{2/n}}} - {{{2/(n+1)}}}) = all interior terms cancel each other, 


           and only two extreme (very first and very last) terms survive


         = 2 - {{{2/(n+1)}}}.



It implies that the limit of the sum  at  n -->oo  is  2 (two, TWO).   <U>ANSWER</U>.
</pre>

Solved.