24no... simple bracket
Prove that 2[\frac {1}{2n^2-1} + \frac {1}{3(2n^2-1)^3}+\frac {1}{5(2n^2-1)^5}+..........]=\frac {1}{n^2} +\frac {1}{2n^4} +\frac {1} {3n^6} +.........=2log_en-log_e(n+1)-log_e(n-1)
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3 Answers
1The series is in the form : 2x + 2x33 + 2x55+.....
which is obtained by expansion of ln(1+x) - ln(1-x) where x = 12n2-1
ln(1 + 12n2-1) - ln(1 - 12n2-1)
= ln(2n22n2-1) - ln(2n2-22n2-1)
= ln(2n22n2-2)
= ln(n2n2-1)
= ln(n2) - ln(n2-1)
= 2ln(n) - ln(n+1)(n-1)
= 2ln(n) - ln(n+1) - ln(n-1)