Suppose that the series ∑∞k=1ak converges. Prove that
limn→∞1nn∑k=1kak=0
I tried to use the definition of convergence of ∑ak but I'm struggling to find N such that given ϵ>0, if n>N, then −ϵ<1n∑nk=1kak<ϵ but I don't know how to connect them
hope somebody can help
Answer
Let sn=a1+⋯+an→a. Then
n∑k=1kak=n∑k=1k(sk−sk−1)=n∑k=1ksk−n−1∑k=1(k+1)sk=nsn−n−1∑k=1sk.
Hence
1nn∑k=1kak=sn−n−1n⋅1n−1n−1∑k=1sk→a−a=0.
We have used the fact that: If bn→b, then so does b1+⋯+bnn.
Note. However, it is not in general true that nan→0.
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