# Proving that the given sequence converges

I have been having trouble proving the following result, which I have written below.

Let $(a_{n=1}^\infty)$ be a sequence of positive real numbers and let $A_n$ = $\sum\limits_{i=1}^n a_i$ Prove that if $\sum\limits_{n=1}^\infty \dfrac{a_n}{A_n}$ < $\infty$, then $\sum\limits_{n=1}^n a_n$ < $\infty$.

I have been given the following hint.

Assume that$\sum\limits_{n=1}^\infty a_n$ = $\infty$. Let 0 < $\epsilon$ < 1/2. Then for a sufficiently large N $\geq$ 2 and all n > N, $\epsilon$ >$\sum\limits_{j=N}^\infty \dfrac{a_j}{A_j}$ $\geq$ $\sum\limits_{j=N}^n \dfrac{a_j}{A_j}$.

I am unsure how to proceed. Thanks for any help in advance.

• Use that $(A_n)$ is increasing. Mar 20, 2016 at 14:44
• @FriedrichPhilipp I see the hint below by Thomas, but I don't see how it is true. Did you have the same idea as Thomas? Mar 20, 2016 at 15:15
• I don't see how the fact that $(A_n)$ is increasing makes the second inequality in the hint true. In fact, I would think that the inequality would be reversed. After all, if the terms of $(A_n)$ are increasing then fixing the denominator of every term and summing over each $a_j$ should be larger than summing over each $a_j$ while simultaneously increasing the denominator of each term. Mar 20, 2016 at 16:06

By the hint, for all $n\ge N$ we have $\frac 1 {A_n}\sum_{j=N}^na_j < \varepsilon$, hence $\sum_{j=N}^na_j < \varepsilon\sum_{j=1}^na_j$. This is equivalent to $(1-\varepsilon)\sum_{j=N}^na_j < \varepsilon\sum_{j=1}^{N-1}a_j$. Therefore, $\sum_{j=N}^na_j < \frac\varepsilon{1-\varepsilon}\sum_{j=1}^{N-1}a_j$ for all $n\ge N$. Letting $n\to\infty$ proves the claim.

You get $$\varepsilon > \sum_{j=N}^n \frac{a_j}{A_j}\ge \frac{1}{A_n}\sum_{j=N}^n a_j$$ for every $n>N$, since $A_j$ is monotonically increasing.

Can you finish the proof with this hint?

• I don't see how that last inequality is true. Mar 20, 2016 at 15:03
• Wouldn't the last inequality be reversed? Mar 20, 2016 at 15:46
• Yes, it's false, you are right. Choose $A_n$ instead of $A_N$. Then you're good. Mar 20, 2016 at 15:49
• @FriedrichPhilipp I think I see it now. Mar 20, 2016 at 16:09
• @shmiggens This is good to hear/read. ;-) Mar 20, 2016 at 16:10

The $\epsilon$ in the hint seems confusing. All we really need is that since $\sum\limits_{n=1}^\infty\frac{a_n}{A_n}$ converges, there is an $N$ so that $\sum\limits_{n=N}^\infty\frac{a_n}{A_n}\le\frac12$. Therefore, since $A_n$ is increasing, for any $m\ge N$, \begin{align} A_m-A_{N-1} &=\sum_{n=N}^ma_n\\ &\le\sum_{n=N}^m\frac{A_m}{A_n}a_n\\ &\le\sum_{n=N}^\infty\frac{a_n}{A_n}A_m\\ &\le\frac12A_m\tag{1} \end{align} $(1)$ is equivalent to $$A_m\le2A_{N-1}\tag{2}$$ Taking the limit of $(2)$ as $m\to\infty$ gives $$\sum_{n=1}^\infty a_n\le2A_{N-1}\tag{3}$$