This cropped up in a research question I'm tackling. I wish to solve the following optimization problem: $$ \text{minimize}\ \sum_{i=1}^\infty f_i \sum_{j=1}^i \sqrt{f_j} \quad\text{subject to}\ \sum_{i=1}^\infty f_i=1\ \text{and}\ f_i \in [0,1]\ \forall i. $$

My attempt was to apply calculus of variation, and adding the normalization constraint with a Lagrange multiplier:
$$
S_\lambda := \sum_{i=1}^\infty f_i \sum_{j=1}^i \sqrt{f_j} + \lambda\left(1-\sum_{i=1}^\infty f_i \right)
$$
and then vary
$$
\frac{\delta S_\lambda}{\delta f_a} = \sum_{j=1}^a \sqrt{f_j} + \frac12 \frac{1}{\sqrt{f_a}} \sum_{i=a}^\infty f_i -\lambda \overset!= 0
\quad\forall a
$$
*(My first question is: can I even apply calculus of variation in this setting? And, if so, is the expression above correct? I asked a colleague to check who couldn't find any mistake, but you know how it is sometimes.)*

Under the assumption that the variation is correct, I then solved the first and last equation via \begin{align} 0= \frac{\delta S_\lambda}{\delta f_1} &= \sqrt{f_1} + \frac12 \frac{1}{\sqrt{f_1}}\underbrace{\sum_{i=1}^\infty f_i}_{=1} - \lambda &\Rightarrow\quad \lambda &= \frac{1+2f_1}{2 \sqrt{f_1}} \\ 0= \lim_{a\rightarrow\infty} \frac{\delta S_\lambda}{\delta f_a} &= - \lambda + \underbrace{\lim_{a\rightarrow\infty} \sum_{i=1}^a \sqrt{f_i}}_{=:A_\infty} + \frac12 \underbrace{\lim_{a\rightarrow\infty}\frac{1}{\sqrt{f_a}} \sum_{i=a}^\infty f_i}_{=:B_\infty} &\Rightarrow\quad \lambda&=A_\infty + B_\infty \end{align} As by the first line we have that $\lambda$ is finite, both limits must exist. The first sum converges when $f_i = o(1/i^2)$, in which case $B_\infty=0$ and $\lambda = A_\infty$.

*But how to proceed? I don't see a good way of e.g. solving this set of equations recursively.*

I'd be grateful for any pointers!

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