limit of $\sum_{j=1}^{n-1}a_j\frac{1}{n-j}$ as $n\to \infty$ Suppose that $\sum_{j=1}^{\infty}a_j$ is convergent, with $a_j\geq 0$. Is it true that
$$\lim_{n\to \infty}S_n=0,\qquad S_n:=\sum_{j=1}^{n-1} \frac{a_j}{n-j}?$$
Attempt (probably wrong) with Fourier:
Define $a_j=0$ for $j\leq 0$.
$$F(x)=\sum_{n=-\infty}^\infty S_ne^{inx}=\sum_{n=1}^{\infty}a_ne^{inx}\sum_{n=1}^{\infty}\frac{1}{n}e^{inx}=\sum_{n=1}^{\infty}a_ne^{inx}(-\log (1-e^{ix})),\qquad x\in (0,2\pi) $$
Since $\begin{cases}1/n, &n>0 \\ 0, & n \leq 0\end{cases}\in L^2(\mathbb{Z})$, then its Fourier transform $\sum_{n=1}^{\infty}\frac{1}{n}e^{inx}\in L^{2}(0,2\pi)\subset L^1(0,2\pi)$.
Since $\sum_{n=1}^{\infty}a_n<\infty$, then $F\in L^1(0,2\pi)$, and $\lim_{n\to \infty}S_n=0$ (Riemann-Lebesgue lemma).
Anyway, even if this is somehow correct I assume this also has an elementary solution.
Also, is it still true if we only require that $\sum_{j=1}^{\infty}\frac{a_j}{j}$ is convergent? It seems so from numerical experiments.
 A: Consider a version of Lebesgue's DCT for counting measure $\mu$ on $\left(\mathbb{N}, 2^{\mathbb{N}} \right)$, which is defined by $\mu(A) = \mathrm{card}{A}$. Integral $\int_{\mathbb{N}} f \mathbb{d} \mu$ of some function $f: \mathbb{N} \to \mathbb{R}$ with respect to this measures equals $\sum_{j=1}^{\infty}f(j)$.
Let's define a sequence of functions $f_n: \mathbb{N} \to \mathbb{R}$ by $f_n(j) = \frac{a_j}{n-j} \cdot {1}\left\{j \leq n-1\right\}$, where ${1}\left\{\; \cdot \;\right\}$ is the indicator function. Observe that $$\int_{\mathbb{N}} f_n \mathbb{d} \mu = \sum_{j=1}^{n-1}\frac{a_j}{n-j} = S_n.$$
Also, we have $\lim_{n\to\infty} f_n(j) = 0$ for all $j \in \mathbb{N}$ (so pointwise limit of $f_n$ is zero function $f \equiv 0$) and $\left|f_n(j)\right| \leq a_j = g(j)$.
As $\int_{\mathbb{N}} g \mathbb{d} \mu = \sum_{j=1}^{\infty} a_n < \infty$, we obtain
$$
\lim_{n\to\infty} S_n = \lim_{n\to\infty} \int_{\mathbb{N}} f_n \mathbb{d} \mu = \int_{\mathbb{N}} f \mathbb{d} \mu = 0
$$
by applying DCT.
A: We will prove that for each $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $S_N < \epsilon$. Because $\sum_{j \geq 1} a_j$ is convergent, so for each $\epsilon > 0$, there exists an integer $n_\epsilon$ such that $\sum_{j \geq n_\epsilon} a_j < \epsilon/2$. Now if $M_\epsilon := \sup\{a_1, a_2, \ldots, a_{n_\epsilon - 1}\}$, then for sufficiently large $n$ we can write:
\begin{align*}
S_n = \sum_{j = 1}^{n-1}\frac{a_j}{n-j} &= \frac{a_1}{n - 1} + \cdots + \frac{a_{n_\epsilon - 1}}{n - (n_\epsilon - 1)} + \frac{a_{n_\epsilon}}{n - n_\epsilon} + \cdots + \frac{a_{n-1}}{1}\\
&\leq \frac{M_\epsilon}{n - n_\epsilon} + \cdots + \frac{M_\epsilon}{n - n_\epsilon} + a_{n_\epsilon} + \cdots + a_{n-1}\\
&\leq \frac{(n_\epsilon - 1)M_\epsilon}{n - n_\epsilon} + \frac{\epsilon}{2}
\end{align*}
Choose $N$ such that $\frac{(n_\epsilon - 1)M_\epsilon}{N - n_\epsilon} < \frac{\epsilon}{2}$. Then $S_N < \epsilon$.
A: Well, let's see. If $\{a_n\}_{n\geq 1}$ has non-negative terms and $\sum_{n\geq 0}a_n$ is convergent, then the radius of convergence of
$$ A(x) = \sum_{n\geq 1} a_n x^n $$
is at least one and we have convergence over $|x|=1$, too. We may notice that
$$ S_n = \sum_{j=1}^{n-1}\frac{a_j}{n-j} $$
is the coefficient of $x^n$ in the product between $A(x)$ and $\sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x)$, hence by Cauchy's theorem
$$ S_n = \frac{1}{2\pi i}\oint_{|z|=\varepsilon}\frac{-A(z)\log(1-z)}{z^{n+1}}\,dz $$
for any $\varepsilon\in(0,1)$.  Picking $\varepsilon=1-\frac{1}{N+1}$ we have that over $|z|=\varepsilon$ the modulus of $A(z)$ is bounded (quite crudely) by $A(1)$ and the modulus of $\log(1-z)$ is bounded by $\log(N+1)$, so
$$ |S_n|\leq \frac{1}{2\pi}\cdot 2\pi\varepsilon \cdot \frac{A(1)\log(N+1)}{\varepsilon^{n+1}} = A(1)\log(N+1)\cdot  \left(1+\frac{1}{N}\right)^n. $$
We still have the freedom to choose $N$ as a function of $n$, but even the optimal choice just gives
$$ S_n = O(\log n). $$
We may reach the target (in)equality $S_n=o(1)$ by changing the integration path, for instance by considering the rectangle with vertices at $-1+\frac{1}{N}\pm\frac{i}{M}$ and $1-\frac{1}{N}\pm\frac{i}{M}$. The integrals on the top and bottom sides almost cancel out and we are left with the estimation of integrals along short segments, one near $z=-1$ and the other one near $z=1$. Picking $N\approx n$ and $M\approx e^n$ this actually leads to $S_n=O\left(\frac{\log n}{n}\right)$.
A: Denote $R_j=\sum_{k\geqslant j}a_k$. Then by summation by parts,
$$
S_n=\sum_{j=1}^{n-1}\frac{R_j-R_{j+1}}{n-j}=
\sum_{\ell=1}^{n-1}\frac{R_\ell}{n-\ell}-\sum_{\ell=2}^n\frac{R_\ell}{n-(\ell-1)}=-R_N+\frac{R_1}{N-1}+\sum_{\ell=2}^{n-1}R_\ell\left(\frac{1}{n-\ell}-\frac{1}{n-(\ell-1)}\right).
$$
The first two terms go to zero by assumption, for the third one, let us write
$$
\left\lvert \sum_{\ell=2}^{n-1}R_\ell\left(\frac{1}{n-\ell}-\frac{1}{n-(\ell-1)}\right)\right\rvert \leqslant R_1\sum_{\ell=2}^{n_0}\left(\frac{1}{n-\ell}-\frac{1}{n-(\ell-1)}\right)+R_{n_0}\sum_{\ell=n_0+1}^{n-1}\left(\frac{1}{n-\ell}-\frac{1}{n-(\ell-1)}\right)\\
\leqslant R_1 \left(\frac{1}{n-2}-\frac{1}{n-(n_0-1)}\right)+R_{n_0}.
$$
