Problem on the limit of a sequence. Assertion: If $\{\alpha_k\}$ is a positive sequence satisfying $\sum \alpha_k =1$, there exists a positive sequence $\{\beta_k\}$ satisfying $\frac{\alpha_k}{\beta_k}\rightarrow 0$ as $k\rightarrow 0$ and $\sum \beta_k=1$.
I cannot show the above assertion.
I want to know how to construct a sequence $\{\beta_k\}$ with the above propeties.
Can someone please help me?
I am stuck. thank you.
Remarks:
$1$. The sequences are infinite.
$2$. Please note that $\alpha_k>0$ and $\beta_k>0$ for all $k=1,2,3,...$.
 A: Let $$\beta_{k}=\sqrt{\sum_{i=k}^{\infty}\alpha_{i}}-\sqrt{\sum_{i=k+1}^{\infty}\alpha_{i}},$$then it is clearly that $$\sum_{k=1}^{\infty}\beta_{k}=\sqrt{\sum_{k=1}^{\infty}\alpha_{k}}=1$$and$$\lim_{n\rightarrow\infty}\dfrac{\alpha_{n}}{\beta_{n}}=\lim_{n\rightarrow\infty}\dfrac{\alpha_{n}(\sqrt{\sum\limits_{i=n}^{\infty}\alpha_{i}}+\sqrt{\sum\limits_{i=n+1}^{\infty}\alpha_{i}})}{\alpha_{n}}=0.$$
A: Just pick a sequence going to zero, and set $\frac{\alpha_k}{\beta_k}$ equal to it. For instance, if we pick the sequence $\frac 1 k$, that gives $\beta_k = k\alpha_k$.
What about the possibility (i.e. certainty) that this gives $\sum \beta_k>1$? Consider $f(n)=\sum_{k=n}^{\infty}\beta_k$. This is a decreasing function of $n$ (the higher $n$ is, the more terms we drop). So all you need to do is prove that there is some $n$ such that $f(n)<1$. Then fill all the terms of $\{\beta_k\}$ before $n$ with values such that the total sum $\sum \beta$ is equal to $1$. This won't affect the limit of $\frac{\alpha_k}{\beta_k}$, as that doesn't care about any finite subset of terms.
