4

Suppose $a_n>0$ and $$S_0=1,S_n=\sum_{k=1}^{n}a_k.$$ We have the result $\sum a_n$ converges iff $\sum\frac{a_n}{S_n}$ conveges. And also we can show that $$\sum_{n=1}^{\infty}a_n\ \mbox{converges}\iff \sum_{n=1}^{\infty}\frac{a_n}{S_{n-1}}\ \mbox{converges}.$$

$\textbf{Proof:}$ If $\sum_{n=1}^{\infty}a_n$ converges, then $S_n\to S>0$, and $$\frac{a_n}{S_{n-1}}\sim\frac{a_n}{S},\ \ n\to\infty.$$ So $\sum_{n=1}^{\infty}\frac{a_n}{S_{n-1}}$ converges.

If $\sum_{n=1}^{\infty}a_n$ diverges, then $S_n\to \infty$, and $$\frac{a_n}{S_{n-1}}>\log\left(1+\frac{a_n}{S_{n-1}}\right) =\log S_n-\log S_{n-1},$$ this implies $$\sum_{k=1}^{n}\frac{a_k}{S_{k-1}}>\log S_{n+1}\to\infty, n\to\infty.$$ So $\sum_{n=1}^{\infty}\frac{a_n}{S_{n-1}}$ diverges.

Combining the above two result, we can conclude that $$\color{red}{\sum\frac{a_n}{S_{n-1}}\ \mbox{converges}\iff \sum\frac{a_n}{S_n}\ \mbox{converges}}.$$

$\textbf{My question is that is there a direct method to prove:}$ $$\color{red}{\sum\frac{a_n}{S_{n-1}}\ \mbox{converges}\iff \sum\frac{a_n}{S_n}\ \mbox{converges}}.$$ One direction is very easy due to $$\frac{a_n}{S_n}<\frac{a_n}{S_{n-1}}.$$

It is expected to give a direct proof of $$\sum\frac{a_n}{S_n}\ \mbox{converges}\implies \sum\frac{a_n}{S_{n-1}}\ \mbox{converges}$$

Any help and hints will welcome.

Is the following generalization is right? For fixed positive integer $k$, $$\color{red}{\sum_{n\geq k}\frac{a_n}{S_{n-k}}\ \mbox{converges}\iff \sum\frac{a_n}{S_n}\ \mbox{converges}}.$$

Riemann
  • 7,203
  • It does not affect your statement, but I would consider $S_0 = 0$ a more natural choice, instead of $S_0 = 1$. – Martin R May 01 '21 at 08:18

1 Answers1

2

As you already noticed, the implication $$ \sum_{n> k}\frac{a_n}{S_{n-k}}\ \mbox{converges}\implies \sum_{n \ge 1}\frac{a_n}{S_n}\ \mbox{converges} $$ follows immediately from the estimate $\frac{a_n}{S_n} \le \frac{a_n}{S_{n-k}}$.

For the other direction we can mimic the proof in the answer that you referred to. If $\sum_{n \ge 1}\frac{a_n}{S_n}$ converges then there is an index $N$ such that for $q > p \ge N$ $$ \frac 12 \ge \sum_{n=p+1}^q \frac{a_n}{S_n} \ge \sum_{n=p+1}^q \frac{a_n}{S_q} = \frac{S_q-S_p}{S_q} = 1 - \frac{S_p}{S_q} $$ and therefore $ S_q \le 2 S_p$. It follows that $$ \frac{a_n}{S_{n-k}} \le 2 \frac{a_n}{S_{n}} $$ for $n \ge N+k$, so that $\sum_{n> k}\frac{a_n}{S_{n-k}}$ converges.

This proves the conjectured generalisation. It is a “direct” proof in the sense that we did not work with the convergence or divergence of $\sum a_n$.

Martin R
  • 113,040