Prove there exists a sequence $\{y_n\}\subset (0,\infty)$ such that $\bar{\lim_{n\rightarrow \infty}}y_n = \infty$

Question

Let $\{a_n\}\subset \mathbb{R}$ a sequence of positive numbers such that the serie $\sum_{n=1} a_n $converge

Prove exists a sequence $\{y_n\}\subset (0,\infty)$ such that

$$\bar{\lim_{n\rightarrow \infty}}y_n = \infty$$

and

$$\sum_{n=1}a_n y_n$$ converge.

My attempt:<\b> Suppose by contradiction that forall sequence $\{y_n\}\subset (0,\infty)$ then

$$\bar{\lim_{n\rightarrow \infty}}y_n \not = \infty$$

or

$$\sum_{n=1}a_n y_n$$ diverge.

Let $\{y_n\}=\{a_n\}$ then if $\liminf_{n\rightarrow \infty}a_n=\limsup_{n\rightarrow \infty}a_n = L $ and $L\not = 0$ then $\sum_{n=1} a_n $

Moreover if $\liminf_{n\rightarrow \infty}a_n\not =\limsup_{n\rightarrow \infty}a_n$ then $\lim a_n$ not exists and for that $\sum_{n=1} a_n $ diverge.

For $$\sum_{n=1}a_n y_n$$ diverge. consider $y_n=1$ for all n and this implies $\sum_{n=1} a_n $ diverge.

Answer 1

Solution: y_n = 2 ^ k for n = n_k = min{n: a_n <= 1 / 2 ^ (2 * k)} (only for n = n_k s.t. n_k < n_(k+1)), and y_n = 1 / 2 ^ n otherwise.

Answer 2

You can't prove by contradiction for all sequences $\{y_n\}\subset (0,\infty)$ by just proving it for specific ones. As such, I don't see how to finish using what you started.

Instead, I would approach this by trying to show how to construct a sequence $\{y_n\}$ which satisfies the given conditions.

Since the series $\sum_{n=1} a_n$ converges, this means that

$$\sum_{n=1}^{\infty}a_n = L \tag{1}\label{eq1A}$$

for some $L \gt 0$. Since $a_n$ are all positive, this means the partial sums, i.e.,

$$\sum_{n=1}^{m}a_n = L_m \tag{2}\label{eq2A}$$

are a strictly increasing sequence as $m$ increases. Have $n_0 = 0$ and for each integer $k \ge 1$, have $n_k$ be the largest index $m$ where $L_m \lt \frac{(2^k - 1)L}{2^k}$. Set

$$y_i = \frac{(1.5)^{k-1}}{L} \text{ for } n_{k-1} \lt i \le n_k \tag{3}\label{eq3A}$$

The sum of the $a_i$ items between this partial sum boundary, i.e., $\frac{(2^k - 1)L}{2^k}$ and the previous one, i.e., $\frac{(2^{k-1} - 1)L}{2^{k-1}}$, would be less than double the difference, i.e., you'll have that sum less than as shown below

$$\begin{equation}\begin{aligned} \sum_{n=n_{k-1} +1}^{n_k}a_n & \lt 2\left(\frac{(2^k - 1)L}{2^k} - \frac{(2^{k-1} - 1)L}{2^{k-1}}\right) \\ & = 2L\left(\frac{2^k - 1}{2^k} - \frac{2^{k} - 2}{2^{k}}\right) \\ & = 2L\left(\frac{2^k - 1 - 2^k + 2}{2^k}\right) \\ & = \frac{L}{2^{k-1}} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Combining \eqref{eq3A} with \eqref{eq4A} gives

$$\begin{equation}\begin{aligned} \sum_{n=n_{k-1} +1}^{n_k}a_n y_n & \lt \left(\frac{(1.5)^{k-1}}{L}\right)\left(\frac{L}{2^{k-1}}\right) \\ & = (0.75)^{k-1} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

Note that, apart from the set of values up to $n_1$, adding the set of values comprising each partial sum from \eqref{eq5A} for $k = 1$ to infinity shows that

$$\begin{equation}\begin{aligned} \sum_{n=1}^{\infty}a_n y_n & \lt \sum_{k=0}^{\infty}(0.75)^k \\ & = 4 \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

Also, from \eqref{eq3A}, you can see that $\{y_i\}$ is an increasing sequence without any upper bound, i.e.,

$$\lim_{n \to \infty} y_n = \infty \tag{7}\label{eq7A}$$