Harmonic Series and Its Divergence by Abel Sum and Cesaro Sum

Question

Already I know that harmonic series, $$\sum_{k=1}^n\frac1k $$ is divergent series.

And, it is also divergent by Abel Sum or Cesaro Sum.

However, I do not know how to prove it is divergent by concept of Abel or Cesaro.

Abel Sum or Cesaro Sum do not exist in this problem.

But, how can I prove it?

I have tried to prove for long time...

Can you please show your work?

Answer 1

A simple trick is to consider that: $$\frac{1}{n}\geq\log\left(1+\frac{1}{n}\right)\tag{1}$$ holds regardless of which concept of convergence you are using, and the partial sums of the RHS of $(1)$ are fairly easy to compute by the telescopic property: $$ \sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\log\prod_{n=1}^{N}\frac{n+1}{n} = \log(N+1).\tag{2}$$ As an alternative, given that $H_n=\sum_{k=1}^{n}\frac{1}{k}$, we have: $$ \frac{H_1+\ldots+H_n}{n}=\frac{1}{n}\sum_{j=1}^{n}\sum_{k=1}^{j}\frac{1}{k}=\frac{1}{n}\sum_{j=1}^{n}\frac{n+1-j}{j}=\left(1+\frac{1}{n}\right)H_n-1.\tag{3}$$

Answer 2

Since the term of the Harmonic series are of O(1/n). Suppose, if the Harmonic series is Cesaro summable, then by Hardy's theorem, the series is convergent, but Harmonic series is divergent. So, Harmonic series is not Cesaro summable.