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Well, I was confronted with this wonderful problem:

Given that $\{a_n\}$ is a positive sequence such that $\sum\limits_{n=1}^\infty \dfrac1{a_n}<\infty$. Let $S_n=\sum\limits_{k=1}^na_k$, then prove for any $0\leqslant\alpha\leqslant2$, $$ \sum_{n=1}^{\infty}{\left( \frac{na_n}{S_n} \right) ^{\alpha}\frac{1}{a_n}}<\infty. $$

Moreover, my friend guesses that the following inequality may be correct:

$$ \sum_{n=1}^{\infty}{\left( \frac{na_n}{S_n} \right) ^{\alpha}\frac{1}{a_n}}\leqslant2^{\alpha}\sum_{n=1}^{\infty}{\frac{1}{a_n}}. $$

The constant $2^\alpha$ is the best choice. For $\alpha=0$ and $\alpha=1$, see A version of Hardy's inequality involving reciprocals., which is not hard to prove. But for $\alpha\in\mathbb R_+$, I do not know how to deal with it. Any advice will be appreciated. Thank you!

P.S. The equivalent inequality of integral is this:

Given $f(x)\geqslant0$. Let $F(x)=\int_0^xf(t)\,\mathrm dt$, prove that $$ \int_0^{\infty}{\left( \frac{xf\left( x \right)}{F\left( x \right)} \right) ^{\alpha}\frac{\text{d}x}{f\left( x \right)}}\leqslant 2^{\alpha}\int_0^{\infty}{\frac{\text{d}x}{f\left( x \right)}}. \quad 0\leqslant α\leqslant 2 $$

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Inspired by Aryabhata's answer, it will be proved that$$ \frac{α 2^α}{4} \cdot \frac{n^2}{S_n} + \sum_{k = 1}^n \left( \frac{k a_k}{S_k} \right)^α \frac{1}{a_k} \leqslant 2^α \sum_{k = 1}^n \frac{1}{a_k}. $$

Step 1: If $f$ is a convex function on $[a, b]$, then for any $a \leqslant x_0 < x_1 < x_2 \leqslant b$,$$ \frac{f(x_1) - f(x_0)}{x_1 - x_0} \leqslant \frac{f(x_2) - f(x_0)}{x_2 - x_0}. \tag{1.1} $$

Proof: \begin{align*} (1.1) &\Longleftrightarrow (x_2 - x_0)(f(x_1) - f(x_0)) \leqslant (x_1 - x_0)(f(x_2) - f(x_0))\\ &\Longleftrightarrow (x_1 - x_0) f(x_2) + (x_2 - x_1) f(x_1) \geqslant (x_2 - x_0) f(x_1)\\ &\Longleftrightarrow \frac{x_1 - x_0}{x_2 - x_0} f(x_2) + \frac{x_2 - x_1}{x_2 - x_0} f(x_0) \geqslant f(x_1). \end{align*} Because $f$ is convex on $(a, b)$, then$$ \frac{x_1 - x_0}{x_2 - x_0} f(x_2) + \frac{x_2 - x_1}{x_2 - x_0} f(x_0) \geqslant f\left( \frac{x_1 - x_0}{x_2 - x_0} x_2 + \frac{x_2 - x_1}{x_2 - x_0} x_0 \right) = f(x_1). $$

Step 2: For any $0 \leqslant α \leqslant 2$ and $n, t > 0$, there is$$ \frac{α 2^α}{4} \cdot \frac{n^2}{t + 1} + 2^α t \geqslant \frac{α 2^α}{4} (2n - 1) + \frac{n^α t}{(t + 1)^α}. \tag{2.1} $$

Proof: Define $s = \dfrac{n}{t + 1}$, then\begin{align*} (2.1) &\Longleftrightarrow \frac{α 2^α}{4} \cdot ns + 2^α \left( \frac{n}{s} - 1 \right) \geqslant \frac{α 2^α}{4} (2n - 1) + s^α \left( \frac{n}{s} - 1 \right)\\ &\Longleftrightarrow (s^α - 2^α) \left( \frac{n}{s} - 1 \right) \leqslant \frac{α 2^α}{4} (ns - (2n - 1))\\ &\Longleftrightarrow \left( \left( \frac{s}{2} \right)^α - 1 \right) (n - s) \leqslant \frac{α}{4} (ns - (2n - 1))s\\ &\Longleftrightarrow \frac{\left( \dfrac{s}{2} \right)^α - 1}{α} \leqslant \frac{(ns - (2n - 1))s}{4(n - s)}. \end{align*}

Note that $f(x) = \left( \dfrac{s}{2} \right)^x$ is convex on $[0, +\infty)$, from (1.1) there is$$ \frac{\left( \dfrac{s}{2} \right)^α - 1}{α - 0} \leqslant \frac{\left( \dfrac{s}{2} \right)^2 - 1}{2 - 0}, $$ thus it suffices to prove (2.1) for $α = 2$, i.e.$$ \frac{2n^2}{t + 1} + 4t \geqslant 2(2n - 1) + \frac{n^2 t}{(t + 1)^2}. \tag{2.2} $$

Now,\begin{align*} (2.2) &\Longleftrightarrow 2n^2 (t + 1) + 4t(t + 1)^2 \geqslant 2(2n - 1)(t + 1)^2 + n^2 t\\ &\Longleftrightarrow (t + 2)^2 n^2 - 4(t + 1)^2 n + 2(2t + 1)(t + 1)^2 \geqslant 0. \end{align*} Because the discriminant with respect to $n$ is$$ (-4(t + 1)^2)^2 - 4(t + 2)^2 \cdot 2(2t + 1)(t + 1)^2 = -8t(t + 1)^2 \leqslant 0, $$ then (2.2) holds, which implies (2.1) holds.

Step 3: For any $a_1, \cdots, a_n > 0$ and $0 \leqslant α \leqslant 2$, there is$$ \frac{α 2^α}{4} \cdot \frac{n^2}{S_n} + \sum_{k = 1}^n \left( \frac{k a_k}{S_k} \right)^α \frac{1}{a_k} \leqslant 2^α \sum_{k = 1}^n \frac{1}{a_k}, \tag{3.1} $$ where $S_k = \sum\limits_{j = 1}^k a_j$.

Proof: It will be proved by induction on $n$. For the base case $n = 1$, because from (1.1) there is$$ \frac{\left( \dfrac{1}{2} \right)^α - 1}{α - 0} \leqslant \frac{\left( \dfrac{1}{2} \right)^2 - 1}{2 - 0} = -\frac{3}{8} < -\frac{1}{4} \Longrightarrow \frac{α 2^α}{4} + 1 \leqslant 2^α, $$ then$$ \frac{α 2^α}{4} \cdot \frac{1}{S_1} + \left( \frac{a_1}{S_1} \right)^α \frac{1}{a_1} = \left( \frac{α 2^α}{4} + 1 \right) \frac{1}{a_1} \leqslant 2^α \cdot \frac{1}{a_1}. $$

Now assume (3.1) holds for $n - 1$. To prove for $n$, by induction hypothesis it suffices to prove that$$ \frac{α 2^α}{4} \cdot \frac{n^2}{S_n} + \left( \frac{n a_n}{S_n} \right)^α \frac{1}{a_n} - \frac{α 2^α}{4} \cdot \frac{(n - 1)^2}{S_{n - 1}} \leqslant 2^α \cdot \frac{1}{a_n}. \tag{3.2} $$ Define $t = \dfrac{S_{n - 1}}{a_n}$, then\begin{align*} (3.2) &\Longleftrightarrow \frac{α 2^α}{4} \cdot \frac{n^2}{t + 1} + \left( \frac{n}{t + 1} \right)^α - \frac{α 2^α}{4} \cdot \frac{(n - 1)^2}{t} \leqslant 2^α\\ &\Longleftrightarrow \frac{α 2^α}{4} n^2 \cdot \frac{t}{t + 1} + \frac{n^α t}{(t + 1)^α} - \frac{α 2^α}{4} (n - 1)^2 \leqslant 2^α t\\ &\Longleftrightarrow \frac{α 2^α}{4} n^2 + \frac{n^α t}{(t + 1)^α} - \frac{α 2^α}{4} (n - 1)^2 \leqslant \frac{α 2^α}{4} \cdot \frac{n^2}{t + 1} + 2^α t\\ &\Longleftrightarrow \frac{α 2^α}{4} (2n - 1) + \frac{n^α t}{(t + 1)^α} \leqslant \frac{α 2^α}{4} \cdot \frac{n^2}{t + 1} + 2^α t, \end{align*} where the last inequality holds by (2.1). End of induction.

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