Since an=∑k=1nn+k1 , then an+1=∑k=1n+1(n+1)+k1 and an+1−an=2n+11+2n+21−n+11=2n+11−2n+21=(2n+1)(2n+2)1>0 . Hence an+1>an . So , the sequence is increasing.
Because for all k such that 1≤k≤n the inequality (1+n)≤(k+n) is true and k+n1≤1+n1 . Therefore, 0<an≤∑k=1nn+11==nn+11+...+n+11=n+1n<1 .
Thus, it is proved that the increasing sequence is bounded from above, hence the sequence converges.
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