# Prove that if $\sum a_n$ converges/diverges so does $\sum {a_n/s_n}$ converge/diverge accordingly

There was this question where it was given that $a_n > 0$ and $s_n = a_0 + a_1 + a_2 +..........+a_n$ and here I was to prove that if : $\sum a_n$ converges/diverges so does $\sum {a_n/s_n}$ converge/diverge likewise ;

Now here was how I approached it : consider that I am trying to prove that $\sum {a_n/s_n}$ is converging then say I am trying to prove that $\lim_{n \to \infty} [({a_n/s_n})/({a_{n-1}/s_{n-1}})] <1$ which is equivalent to $\lim_{n \to \infty} (a_n* s_{n-1})/\lim_{n \to \infty} (s_{n}*a_{n-1})$, now since we already know that $a_n$ is converging we can say that $\lim_{n \to \infty} a_n = 0$ thus using this we can say that the ratio $\lim_{n \to \infty} (a_n* s_{n-1})/\lim_{n \to \infty} (s_{n}*a_{n-1}) = 0$ but in the denominator too we have $\lim_{n \to \infty} a_{n-1}$ which too tends to $0$ as $\lim_{n \to \infty} a_n = 0$ and then we have $0/0$ which is absurd, so I abandoned this method.

Can anyone modify by method above or was it really of no great use ?

The second method I approached it was by trying to prove that the ratio $a_n/({a_n/s_n})$ is positive and finite which if proved then $\sum a_n$ and $\sum {a_n/s_n}$ would converge or diverge together. Now if $\sum a_n = s_n$ was converging then it's done but if $\sum a_n$ were diverging then that would really unsettle this method since in $a_n/s_n, s_n = \sum a_n$ and $\sum_{n = o}^{n = \infty} a_n = \infty$, again I would get the ratio $a_n/s_n = 0$ and thus I turn to here.

Was there any mistake in my approach above ? And could it be modified ?or just could anyone suggest me a better approach ?

Lemma: Suppose $0< b_n <1$ for all $n.$ Then $\prod_n b_n > 0 \iff \sum_n (1-b_n) < \infty.$

Suppose $s_n \to \infty.$ Summing over $n>1,$ we have

$$\sum a_n/s_n = \sum (s_n-s_{n-1})/s_n = \sum (1-[s_{n-1}/s_n]).$$

If this sum converges, then by the lemma,

$$\frac{s_1}{s_2} \frac{s_2}{s_3}\cdots \frac{s_{n-1}}{s_n} = \frac{s_1}{s_n} \to L > 0.$$

That's a contradition, so $\sum a_n/s_n = \infty.$

• the lemma you stated above ,just could a cite source where I can look upon it – Arnav Das Oct 26 '15 at 7:36

Note that we have:

$$\frac{a_n}{s_n}=\frac{s_n-s_{n-1}}{s_n}\leq \int_{s_{n-1}}^{s_n}\frac{dt}{t}\leq \frac{s_n-s_{n-1}}{s_{n-1}}=\frac{a_n}{s_{n-1}}$$

1) Suppose first that your series $\displaystyle\sum a_n$ is convergent. Then there exists $L$ such that $s_n\leq L$ for all $n$. By the above, we get for all $N$ that $$\sum_{n=1}^{N}\frac{s_n-s_{n-1}}{s_n} \leq \int_{s_{0}}^{s_N}\frac{dt}{t}\leq \int_{s_0}^{L}\frac{dt}{t}$$

Hence the series $\displaystyle \sum \frac{a_n}{s_n}$ is convergent.

2) Suppose now that the series $\displaystyle \sum a_n$ is divergent. If $\displaystyle \frac{a_n}{s_n}$ does not $\to 0$, the series $\displaystyle \sum \frac{a_n}{s_n}$ is divergent. If $\displaystyle \frac{a_n}{s_n} \to 0$, there exists $K$ such that $n\geq K$ imply $\displaystyle s_n-s_{n-1}=a_n\leq \frac{s_n}{2}$. This imply $s_n\leq 2s_{n-1}$, and then $$\frac{s_n-s_{n-1}}{s_n}\geq \frac{1}{2}\frac{s_n-s_{n-1}}{s_{n-1}}$$ Now use the second inequality with the integral to finish.