## Task 30/67

The sum is sought

$$
\sum_{k=1}^{\infty} a_{k}=a_{1}+a_{2}+a_{3}+\ldots+a_{n}+\ldots
$$

of which the following is known:

1. The terms of the series form a monotonic sequence.
2. It is $a_{1}=1, a_{n}>0$ for every $n$.
3. It is $\prod_{k=1}^{\infty} a_{2 k-1}=\prod_{k=1}^{\infty} a_{2 k}$.
4. It is $a_{2 k}=a_{2 k-2}-a_{2 k-1}$.
From condition 3, it follows that

$$
a_{1} \cdot a_{3} \cdot \ldots \cdot a_{2 k-1} \cdot \ldots = a_{2} \cdot a_{4} \cdot \ldots \cdot a_{2 k} \cdot \ldots \quad \text { or } \quad \frac{a_{1}}{a_{2}} \cdot \frac{a_{3}}{a_{4}} \cdot \ldots \cdot \frac{a_{2 k-1}}{a_{2 k}} \cdot \ldots = 1
$$

If the sequence $\left\{a_{k}\right\}$ were strictly monotonically increasing, then $a_{2 k-1} < a_{2 k}$. Since $a_{2 k-1} > 0$, it also holds that $a_{2 k-2} > a_{2 k-1}$, so the sequence is monotonically decreasing. All $a_{k}$ with $k \geq 3$ are thus less than 1, and

$$
\prod_{k=1}^{\infty} a_{2 k-1} = 0, \quad \prod_{k=1}^{\infty} a_{2 k} = 0 \quad \text { therefore } \quad \prod_{k=1}^{\infty} a_{2 k-1} = \prod_{k=1}^{\infty} a_{2 k}
$$

From conditions 1), 2), and 4) alone, the sum of the sequence cannot be uniquely determined, as $a_{2}$ and all $a_{2 k-1}$ are freely selectable.

If condition 3) of the problem is changed so that

$$
\prod_{k=1}^{\infty} a_{2 k-1} = \prod_{k=1}^{\infty} a_{2 k} \quad \text { is changed to } \quad \prod_{k=1}^{n} a_{2 k-1} = \prod_{k=1}^{n} a_{2 k}
$$

then the problem is uniquely solvable. For $k=1$, it follows that $a_{1} = a_{2}$, for $k=2$, it follows that $a_{1} \cdot a_{3} = a_{2} \cdot a_{4}$, and so on. This implies $a_{2 k-1} = a_{2 k}$.

Since $a_{2 k} = a_{2 k-2} - a_{2 k-1}$, it follows that $a_{2 k} = \frac{1}{2} a_{2 k-2}$. With this, the terms of the sequence can be determined. One obtains:

$$
S = 1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \ldots + \frac{1}{2^{k}} + \frac{1}{2^{k}} + \ldots = 4
$$