Determine the maximum value of the sum $S=\sum_{n=1}^\infty \frac{n}{2^{n}} (a_1a_2…a_n)^{1/n} $

Question

Determine the maximum value of the sum $$S=\sum_{n=1}^\infty \frac{n}{2^{n}} (a_1a_2…a_n)^{1/n} $$ over all sequences $a_k$ of non negative real numbers satisfying $$\sum_{k=1}^\infty a_k =1.$$ I have an idea to solve this that the product may be transformed into sum by using AM-GM inequality? If I take $$\frac{a_1+a_2+…+a_n}{n}\ge (a_1a_2…a_n)^{1/n}$$ I reach to a wrong answer since all $a_k$ can’t be equal together so what other adjustment can be done? In a solution I got, they by default assumed it to be a GP with common ratio $1/4$ and $a_1=3/4$ but I wonder why did they choose this only?

Attempt $2$: If I assume it to be a GP of first term $a_1=a$ common $r$ then I get $\frac{a}{1-r}=1$ and putting this in $S$, I got $$S=a \sum_{n=1}^\infty\frac{n}{2^{n}}(1-a)^{\frac{n-1}{2}}$$ Now how can I proceed (maybe through derivatives but how?)?

I’m not attaching the official solution since I want a method other than that. As mentioned above they take $a=3/4$ and $r=1/4$ arbitrarily to solve but the question is why this pair only?

Answer 1

I have no idea how to prove that the optimal solution is geometric, but if you assume it is, you have for all $n$, $a_n = a_1r^{n - 1}$ for some $0 \leqslant r < 1$ and $\sum_{n = 1}^\infty a_n = \frac{a_1}{1 - r} = 1$ hence $a_1 = 1 - r$. From here, you have for all $n$, $$ \frac{n}{2^n}(a_1\cdots a_n)^{1/n} = (1 - r)\frac{n}{2^n}(r^0\cdots r^{n - 1})^{1/n} = (1 - r)\frac{n}{2^n}r^{(n - 1)/2} = \frac{1 - r}{2}n\left(\frac{\sqrt{r}}{2}\right)^{n - 1}. $$ Now, use the fac that $\sum_{n \geqslant 1} x^n = \frac{1}{1 - x} - 1$ and differentiate this equality to deduce that $\sum_{n \geqslant 1} nx^{n - 1} = \frac{1}{(1 - x)^2}$. Therefore, $$ \sum_{n \geqslant 1} \frac{n}{2^n}(a_1\cdots a_n)^{1/n} = \frac{1 - r}{2}\sum_{n \geqslant 1} n\left(\frac{\sqrt{r}}{2}\right)^{n - 1} = \frac{1 - r}{2}\frac{1}{\left(1 - \frac{\sqrt{r}}{2}\right)^2} = \frac{2 - 2r}{(2 - \sqrt{r})^2}. $$ It means that we must find the $r$ between $0$ and $1$ that maximizes $f(r) = \frac{2 - 2r}{(2 - \sqrt{r})^2}$. We have if $0 < r < 1$, \begin{align*} f'(r) & = \frac{-2 \cdot (2 - \sqrt{r})^2 - (2 - 2r) \cdot 2\frac{-1}{2\sqrt{r}}(2 - \sqrt{r})}{(2 - \sqrt{r})^4}\\ & = \frac{-2\sqrt{r}(2 - \sqrt{r}) + 2 - 2r}{\sqrt{r}(2 - \sqrt{r})^3}\\ & = \frac{2 - 4\sqrt{r}}{\sqrt{r}(2 - \sqrt{r})^3} \end{align*} Therefore, $f'(r)$ vanishes at $r = 1/4$, is positive before and negative after. We deduce that $r = 1/4$ is the optimal value and in this case, the sum is $f(1/4) = \frac{2 - 2\frac{1}{4}}{\left(2 - \sqrt{\frac{1}{4}}\right)^2} = \frac{2 - \frac{1}{2}}{\left(2 - \frac{1}{2}\right)^2} = \frac{2}{3}$.

Answer 2

Let $S = \sum_{n=1}^\infty \frac{n}{2^n}\left( \prod_{i=1}^n a_i \right)^\frac{1}{n} = \sum_{n=1}^\infty\frac{n}{2^n} r^{\frac{i+1}{2}}\left( \prod_{i=1}^n \frac{a_i}{r^i} \right)^\frac{1}{n} \leq \sum_{n=1}^\infty\frac{1}{2^n} r^{\frac{n+1}{2}}\left( \sum_{i=1}^n\frac{a_i}{r^i} \right) = $ $= \sum_{i=1}^\infty \sum_{n=i}^\infty \frac{1}{2^n} r^{\frac{n+1}{2}} \frac{a_i}{r^i} $

Now $\sum_{n=i}^\infty \frac{1}{2^n} r^{\frac{n+1}{2}} \frac{a_i}{r^i} = \frac{a_i}{\left(4r\right)^\frac{i}{2}} r^\frac{1}{2} \sum_{n=0}^\infty \left(\frac{r^\frac{1}{2}}{2}\right)^n = \frac{a_ir^{\frac{1}{2}}}{(4r)^{\frac{i}{2}}} \frac{1}{1-\frac{r^{\frac{1}{2}}}{2}} = C\frac{a_i}{(4r)^\frac{i}{2}} $ In order to get useful information, $4r=1$, because that way we can use the bound on the sum of $a_i$.

Answer 3

What @Jorge wrote should be very helpful to see why we choose $r=1/4$. I will try to explain in more depth the complications your initial approach had and the motivation for choosing a geometric sequence.

Your Initial Approach

As you noted, if we try to attack the problem immediately with AM-GM, we get the following:

$$\sum_{n=1}^\infty \frac{n}{2^n}\prod_{j=1}^na_j^{1/n} \le \sum_{n=1}^\infty \sum_{j=1}^n \frac{n}{2^n}a_j = \sum_{j \ge 1}\sum_{n\ge j} \frac{1}{2^n}a_j = \sum_{j\ge 1} 2^{1-j}a_j.(1)$$

However, there isn't much we can do here since all we know is $\sum_{j\ge 1 } a_j = 1$.

Motivation for Steps Forward

We should strive to manipulate the expression $\frac{n}{2^n}\prod_{j=1}^na_j^{1/n}$ such that when AM-GM is applied, and we flip the sum for simplification $\sum_{j\ge 0} a_j$ pops out.

If you play with the expression some before applying AM-GM, multiplying and dividing out numbers to the $a_j$'s, you'll see that you can affect the resulting coefficients of the $a_j$'s in your upper bound. So let's try to play with the coefficients in a very controlled manner using a sequence, say $\{r_j\}_{j \ge 1}$:

$$\sum_{n=1}^\infty \frac{n}{2^n\prod_{k=1}^n r_k^{1/n}} \prod_{j=1}^n (a_jr_j)^{1/n} \le \sum_{n=1}^\infty \frac{1}{2^n\prod_{k=1}^n r_k^{1/n}}\sum_{j=1}^n a_jr_j = \sum_{n=1}^\infty\sum_{j=1}^{n} \frac{a_jr_j}{2^n\prod_{k=1}^nr_k^{1/n}} $$.

$$=\sum_{j=1}^\infty\sum_{n=j}^{\infty} \frac{a_jr_j}{2^n\prod_{k=1}^nr_k^{1/n}} = \sum_{j=1}^\infty a_j r_j\sum_{n=j}^{\infty} \frac{1}{2^n\prod_{k=1}^nr_k^{1/n}}.(2)$$

At this point this point, the only hope is if we consider $\{r_k\}_{k \ge 1}$ as a geometric sequence; otherwise the calculation seems intractable. Additionally, we're hinted towards this direction since we'd like the sum in $n$ to be a constant.

Following Through on a Hope

(After this point you could refer to Jorge's work, it is fundamentally the same.)

So let $r_k = r^k$:

$$\sum_{j=1}^\infty a_j r^{j-1/2}\sum_{n=j}^{\infty} \frac{1}{2^n r^{\frac{n}{2}}} = \sum_{j=1}^\infty a_j r^{j-1/2}\frac{1}{2^jr^{j/2}} \sum_{n=0}^\infty \frac{1}{(2r^{1/2})^n} = \sum_{j=1}^\infty a_j\frac{r^{j-1/2}}{2^jr^{j/2}} \frac{2r^{1/2}}{2r^{1/2}-1} = \sum_{j=1}^\infty a_j\frac{ r^{j/2}}{2^{j-1}} \frac{1}{2r^{1/2}-1}= \frac{1}{2r^{1/2}-1}\sum_{j=1}^\infty a_j \frac{ r^{j/2}}{2^{j-1}}$$.

Remember that we want $\sum_{j\ge1}a_j$ up to some constant so this means that $r^{j/2}/2^{j-1}$ needs to be a constant. We note that $r^{1/2}/1 = c $ and $r/2 = c $; using the ratio of these two equalities, we get $\Rightarrow r^{1/2} = 2 \Rightarrow r = 4$.

This gives us that the sum should be bounded by $\frac{1}{4-1} (\sum_{j \ge 1} a_j ) \frac{4^{j/2}}{2^{j-1}} = \frac{1}{3} (1) \frac{2^j}{2^{j-1}} = \frac{2}{3}$.

The other half of the problem is to find a sequence which meets the upper bound. Once that is done you've shown it's the maximum.

Takeaway

My recommendation is that you write out the expanded form in columns and rows of the sum your initial approach gave you (1). Then expand the sum I give in (2). Then, if you stare at (2) expanded long enough, you'll get the impression that the problem is sort of analogous to a double-counting problem (except instead of things being overcounted, they're being undercounted); so perhaps we could use the sequence $\{r_j\}_{j \ge 1}$ to redress this undercounting.