∞∑n=1log(1+an) converges absolutely⇔∞∑n=1an converges absolutely.
How to prove this,
Suppose ∞∑n=1an converges absolutely.
Let un=an and vn=log(1+an), then limn→∞unvn=1>0⟹∞∑n=1log(1+an) converges absolutely.
How to prove the converse part?
Answer
Hint: From the definition of ln′(1), we have
limu→0ln(1+u)u=1.
Thus there is a>0 such that
12≤|ln(1+u)u|≤32
for u∈(−a,a),u≠0.
No comments:
Post a Comment