Does the following series converges or diverges? $$\sum_{n=1}^\infty\frac{1}{n(1+\frac{1}{2}+\cdots+\frac{1}{n})}$$ Myattemp:
let $a_n=(\frac{1}{n})$ and $b_n=(\frac{1}{1+\frac{1}{2}+\cdots+\frac{1}{n}})$, It is enough to apply Dirichlet's test
It follows that$$\sum_{n=1}^\infty\frac{1}{n(1+\frac{1}{2}+\cdots+\frac{1}{n})}$$ is convergent.
The series diverges. By partial summation, you can establish $$\sum_{k\le n} \frac1k = \log n + O(1)$$ i.e., for some $n_0 \in \Bbb N$ and $C_0 > 0$, we have $\sum_{k\le n} \frac1 k \le \log n + C_0$ for every $n\ge n_0$. Next, you can find a constant $C_1 > 0$, and $n_1 \in \Bbb N$ such that $\log n + C_0 \le C_1 \log n$ for every $n \ge n_1$. Choose $m = \max\{n_0, n_1\}$. Thus, $\sum_{k\le n} \frac 1 k \le C_1 \log n$ for all $n\ge m$.
Finally, $$\sum_{n=1}^\infty\frac{1}{n(1+\frac{1}{2}+\cdots+\frac{1}{n})} \ge \sum_{n=m}^\infty\frac{1}{n(1+\frac{1}{2}+\cdots+\frac{1}{n})} \ge \frac{1}{C_1} \sum_{n=m}^\infty \frac{1}{n\log n} = \infty$$ as the series $\sum_{n=m}^\infty \frac{1}{n\log n}$ diverges by the Cauchy condensation test, for example.
In Dirichlet's Test, $b_n$ must be bounded by a constant. But, since in your case, $b_n = H_n$, the $n$-th harmonic number, and the fact that $H_n \to \infty$ means the conditions of Dirichlet's test do not apply here.
In fact, the sum is divergent. Can you use what you know about $H_n$ to prove the sum is divergent?
Hint: $\text{D}{(\log(\log(n)))} = 1/(n\log n)$
