Question from a first course in real analysis: Suppose $a_n$ is a sequence of non negative reals
Suppose further that for any sequence $b_n$ that is non negative and converges to $0$, $\sum a_n b_n$ converges
Prove that $\sum a_n$ converges
I asked about this online and someone suggested summation by parts could help, but that is not on the course so I don’t think it is. Unless they expect us to derive summation by parts on our own. I don’t even know what summation by parts is.
This is a proof that uses Summation by parts in the following form.
Let $ f:\mathbb{R} \to \mathbb{R} $ be a continuously differentiable function on the interval $ [x,y] $. $($Here $n$ runs through the positive integers and $x,y \in \mathbb{R}$ are arbitrary.$)$ Let
\begin{equation}
A(t):=\sum_{n\leq t}a_{n}. \\\nonumber
\end{equation}
Then
\begin{equation}
\sum_{x < n \leq y} a_{n}f(n) = A(y)f(y) - A(x)f(x) - \int_{x}^{y} A(t)f'(t)\,dt. \\\nonumber
\end{equation}
We want to show that for any sequence $a_n$, for which $\sum a_n$ diverges (in our case goes to $\infty$, since $a_n$ is non-negative), there exists a non-negative sequence $b_n$, for which $\sum a_nb_n$ also diverges. Let $$b_n=\frac{1}{|\log(A(n))|}\text{, for }A(n)\neq1\text{ and }b_n=0\text{ for }A(n)=1.$$ Note that there exists $k \in \mathbb{N}$, such that $b_n$ is decreasing for $n\geq k$, since $A(n)$ is increasing and goes to infinity (it's bigger than one from some point on). We can also extend to a differentiable decreasing $B(n)$, such that $B(n)=b_{n}$ for all $n\in\mathbb{N}$ (for $n\geq k $). Next, we're going to show that the following sum is divergent, using summation by parts. $$\sum_{n=k}^{\infty}a_{n}b_{n}=\sum_{n=k}^{\infty}a_{n}B(n)=\lim_{c\to\infty}A(c)B(c)-a_{k}B(k)-\int_{k}^{\infty}A(n)B'(n)\,dn.$$Now, $a_kB(k)$ is finite and the integral is non-negative since $B(n)$ is decreasing for $n\geq k $. Furthermore, $$\lim_{c\to\infty}A(c)B(c)=\lim_{c\to\infty}\frac{A(c)}{|\log(A(c))|}$$ and since $\log(x)<\sqrt{x}$ (we can drop the absolute value sign, since $A(n)>1$ from some point), this limit diverges and thus $\sum_{n=k}^{\infty}a_{n}b_{n}$ also diverges. Finally, since $\sum_{1}^{k}a_{n}b_{n}\geq0$ we know that $\sum a_{n}b_{n}$ diverges.
