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Suppose that $a_j$ are non-negative real numbers such that $\sum_{j=1}^\infty a_j<\infty$. Is it true that $$ \sum_{j=1}^nja_j=o(n) $$ as $n\to\infty$?

I am not sure if it is true or not. It is straightforward to show that $\sum_{j=1}^nja_j=O(n)$ as $n\to\infty$. We have that $$ \sum_{j=1}^nja_j\le n\sum_{j=1}^na_j\le n\sum_{j=1}^\infty a_j $$ for each $n\ge 1$. However, the bound is trivial and maybe there is a sharper one.

Any help is much appreciated!

Cm7F7Bb
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    Note that $$\sum_{j=1}^nja_j\leqslant\sum_{j=1}^nA_{j}$$ with $$A_{j}=\sum_{k=j}^\infty a_k$$ and that $A_{j}\to0$ when $j\to\infty$ hence, for every positive $\epsilon$, there exists some $k$ such that for every $j\geqslant k$, $A_{j}\leqslant\epsilon$. For every $n\geqslant k$, $$\sum_{j=1}^nja_j\leqslant kA_{1}+(n-k)\epsilon$$ This implies $$\limsup\frac1n\sum_{j=1}^nja_j\leqslant\epsilon$$ for every positive $\epsilon$ hence the claim holds. – Did Aug 09 '16 at 09:12
  • @Did Thank you very much! Could you post your comment as an answer so that I can upvote and accept it? – Cm7F7Bb Aug 09 '16 at 09:22
  • You could instead post an expanded version of this comment as an answer, and, if users think it is correct, accept it after a while. – Did Aug 09 '16 at 09:30
  • @Did. It's perfectly correct. – DanielWainfleet Aug 10 '16 at 00:46

3 Answers3

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This is a shortened version of the accepted answer. Since (as $n\to\infty$) $$A_n:=\sum_{k\ge n}a_k\to0,$$ so does its Cesàro mean: $$\frac1n\sum_{k=1}^nA_k\to0,$$ hence also: $$\frac1n\sum_{j=1}^nja_j=\left(\frac1n\sum_{k=1}^nA_k\right)-A_{n+1}\to0.$$

Anne Bauval
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Let $A_k=\sum_{j=1}^{k}a_j$. By summation by parts:

$$ \sum_{j=1}^{n}ja_j = (n+1) A_n - \sum_{j=1}^{n} A_j. \tag{1}$$ For any $\varepsilon>0$, let $n_\varepsilon$ the least natural number such that $\forall n\geq n_\varepsilon$, $\left|A_n-\lim_{n\to +\infty} A_n\right|\leq\varepsilon.$
Let $n=k n_\varepsilon$ and $C=\lim_{n\to +\infty}A_n$: by $(1)$ we have $$ \sum_{j=1}^{n} ja_j \leq n\varepsilon+ (n_\varepsilon+1) (C+\varepsilon)+\sum_{j=1}^{n_\varepsilon}A_j \tag{2}$$ hence for any $n$ big enough we have $\sum_{j=1}^{n}ja_j \leq 2\varepsilon n$. Since $\varepsilon$ is arbitrary, the claim follows.

Jack D'Aurizio
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  • I don't understand how you obtain (2). Could you provide some details? Why do we need to assume that $n=kn_\varepsilon$? – Cm7F7Bb Oct 05 '16 at 14:13
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This is Did's comment posted as an expanded answer (this was Did's suggestion).

Let us denote $A_j=\sum_{k=j}^\infty a_k$ and let us observe that $A_j\to0$ as $j\to\infty$ because $\sum_{j=0}^\infty a_j<\infty$. Since $a_j$'s are non-negative real numbers, we have that $$ \sum_{j=1}^nja_j\le\sum_{j=1}^nja_j+nA_{n+1}=\sum_{j=1}^nA_j. $$ For each $\varepsilon>0$, there exists $k(\varepsilon)\ge1$ such that $A_j\le\varepsilon$ for each $j\ge k(\varepsilon)$. For each $n>k(\varepsilon)$, we obtain $$ \sum_{j=1}^nja_j\le\sum_{j=1}^nA_j=\sum_{j=1}^{k(\varepsilon)}A_j+\sum_{j=k(\varepsilon)+1}^nA_j\le k(\varepsilon)A_1+(n-k(\varepsilon))\varepsilon. $$ This implies that $$ \limsup_{n\to\infty}\frac{\sum_{j=1}^nja_j}n\le\varepsilon $$ for each $\varepsilon>0$. Hence, $\sum_{j=1}^nja_j=o(n)$ as $n\to\infty$.

Cm7F7Bb
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