Sunday, 3 July 2016

real analysis - Convergence of the series sumlimitsinftyn=1frac1nlogleft(1+frac1nright).



I am trying to check the convergence or divergence of the series n=11nlog(1+1n).





My attempt: for a finite
p,n+pk=n1klog(1+1k)<1nn+pk=nlog(1+1k)=1nlogΠn+pk=n(k+1k)=1nlog(1+p+1n)<1nlog2, for large n and p is finite.<ε
Hence the series converges.



Answer



Because 1nln(1+1n)1n21 and
n=11n2=π26.



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