Wednesday, 21 October 2015

real analysis - sumlimitsin=1nftylog(1+an) converges absolutely iffsumlimitsin=1nftyan converges absolutely.




n=1log(1+an) converges absolutelyn=1an converges absolutely.





How to prove this,



Suppose n=1an converges absolutely. Let un=an and vn=log(1+an), then lim How to prove the converse part?


Answer



Hint: From the definition of \ln'(1), we have



\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1.



Thus there is a>0 such that




\frac{1}{2}\le \left|\frac{\ln (1+u)}{u}\right| \le \frac{3}{2}



for u\in (-a,a),u\ne0.


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