# Prove the divergence of a particular series, given that another series diverges

Suppose $\{a_i\}_{i\in\mathbb N}$ is an increasing sequence of positive real numbers such that $$\sum_{n=1}^\infty\frac{1}{a_n}=+\infty.\tag{1}$$ Then I have to show that also $$\sum_{n=2}^\infty\frac{1}{na_n-(n-1)a_{n-1}}=+\infty.$$ How should I proceed?

EDIT: At first I didn't want to post this problem because I thought it could have been far too easy for you. Then, after seeing this I changed my mind and I hope somebody could have pointed out something as tricky as in that case. This problem is driving me crazy because it seems that every approach I try leads to a dead end.

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Ok I think I came up with a solution for this one.

First of all let us remark the following

Cauchy Condensation Test: For a positive non increasing sequence $\{f(n)\}_n$ the series $$\sum_{n=1}^{\infty}f(n)$$ is convergent if and only if the series $$\sum_{n=0}^\infty 2^nf(2^n)$$ does. Moreover one has in this case $$\sum_{n=0}^{\infty}f(n)\leq\sum_{n=0}^\infty2^nf(2^n)\leq 2\sum_{n=0}^{\infty}f(n).\tag{1}$$

Then observe that $$\left\{\frac{1}{a_i}\right\}_{i\in\mathbb N}$$ is positive and non increasing, therefore the condensation test applies.

Now observe the following

\begin{align}&\frac{2^ka_{2^k}}{2^{k-1}}\geq\frac{2^ka_{2^k}-2^{k-1}a_{2^{k-1}}}{2^{k-1}}\\&=\frac{2^ka_{2^k}-(2^k-1)a_{2^k-1}+(2^k-1)a_{2^k-1}-\dotso+(2^{k-1}+1)a_{2^{k-1}+1}-2^{k-1}a_{2^{k-1}}}{2^{k-1}}.\tag{2}\end{align}

Therefore, using arithmetic-harmonic mean inequality on $(2)$ we can write, for any $k\geq 1$, $$\frac{2^ka_{2^k}}{2^{k-1}}\geq\frac{2^{k-1}}{\sum_{n=2^{k-1}+1}^{2^k}\frac{1}{na_n-(n-1)a_{n-1}}},\tag{3}$$ which in turn implies $$\sum_{n=2^{k-1}+1}^{2^k}\frac{1}{na_n-(n-1)a_{n-1}}\geq \frac{2^k}{4a_{2^k}}.\tag{4}$$ Finishing from here is easy: define indeed $$S_N:=\sum_{n=2}^N\frac{1}{na_n-(n-1)a_{n-1}}.\tag{5}$$ If $N=2^j$ then by $(4)$ we obtain that $$S_{2^j}=\sum_{k=1}^j\sum_{n=2^{k-1}+1}^{2^{k}}\frac{1}{na_n-(n-1)a_{n-1}}\geq \sum_{k=1}^j\frac{2^k}{4a_{2^k}}.\tag{6}$$ By the Cauchy condensation test we have then $$\lim_{j\to+\infty}S_{2^j}=+\infty.\tag{7}$$ Since $S_N$ is increasing in $N$, consider that for a general $N>2$ (i.e. not necessarily of the form $N=2^j$) there exists exactly one natural $q>0$ such that $2^q\leq N<2^{q+1}$. Therefore $$S_N>S_{2^q},\tag{8}$$ from which $$\lim_{N\to+\infty}S_N=+\infty.$$ The proof is complete and yes, that was tricky as well as the other one. Hope as usual to be correct. Cheers.

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