SUM the following series, with every other successive denominator in the bracket being greater than the previous one by "1" ?

#1/3[(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+....]#

1 Answer
Shivang Madaan
Jul 28, 2018

#13/36#

Explanation:

first of all, divide the sum into to parts- the positive constants and the negative constants.
#=1/3{[1/2+1/3+1/4+1/5+......]-[1/5+1/6+1/7+......]}#
cancel the constants of opposite signs,i.e., from #1/5 # to infinity.
#=1/3{1/2+1/3+1/4}#
take LCM on the inside
#=1/3{[6+4+3]/12}#
#=13/36#

hope this helps,
Shivang M.

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