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Question Number 136758 by mathlove last updated on 25/Mar/21
1==>limx→∞(1π+1π2+1π3+⋅⋅⋅+1πn)=? 2=>limx→∞1+22+32+⋅⋅⋅⋅+n21−n3=?
Answered by Dwaipayan Shikari last updated on 25/Mar/21
(1)1π+1π2+...=1π1−1π=1π−1 (2)limn→∞1+22+32+..+n21−n3=n(n+1)(2n+1)6(1−n3)∼−n(n+1)(2n+1)6n3 =−(1+1n)(2+1n)6=−13
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