I ask for some help with this question:
Prove or provide counter example:
If ∞∑n=1nan converges then ∞∑n=1nan+1 also converges.
I tries this way:
If ∞∑n=1nan converges then nan→0, therefore an→0.
There are 3 possible cases:
1) If an>0 and an is monotonic decreasing sequence then $na_{n+1}
2) If an>0 and an is not monotonic decreasing sequence : it is not possible that an+1>an because in this case an→∞, therefore it must be an+1≤an and ∑∞n=1nan+1 converges by Comparison Test.
3) If an is sign-alternating series. There I have a problem to find a solution.
Thanks.
Answer
Yes. Put bn=nan, so the question is now (see my comment on the question):
If ∞∑n=1bn converges, does ∞∑n=1bnn converge?
Let sn=∑nk=1bn. We get (partial summation)
n∑k=1bkk=n∑k=1sk−sk−1k=n∑k=1(1k−1k+1)sk+snn+1=n∑k=11k(k+1)sk+snn+1
which converges as n→∞, because sk is bounded, so the sum is absolutely convergent.
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