I recently found the following problem in my Google+ account: What is

\(\dfrac{1}{2 \cdot 4} + \dfrac{1 \cdot 3}{2 \cdot 4 \cdot 6} + \dfrac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8} + \ldots\)

This kind of sum of growing products can often be solved with the following trick.

\(1 = a_1 + (1-a_1) = a_1 + (1-a_1) a_2 + (1-a_1)(1-a_2) = \ldots\)

The n-th iteration of this process has yields

\(1 = a_1 + \ldots + (1-a_1)\ldots(1-a_n)a_{n+1} + (1-a_1)\ldots(1-a_{n+1}).\)

This is true for any sequence a_n. If we want to let n go to infinity we need to control the last term in this sum. We want to know its limit. E.g.,

\(1 = \sum\limits_{k=1}^\infty \left( a_{k+1} \prod\limits_{l=1}^k (1-a_l) \right)\)

will be true only if

\(\lim\limits_{k \to \infty} \prod\limits_{l=1}^k (1-a_l)  = 0.\)

We can use

\(\log(1-h) \le -h \qquad\text{for } 0 \le h < 1\)

and get the following estimate

\(\log \prod\limits_{l=1}^k (1-a_k) = \sum\limits_{l=1}^k \log (1-a_l) \le – \sum\limits_{l=1}^k a_l.\)

Thus for positive a(l), not larger than 1,

\(\sum\limits_{k=1}^\infty a_l = \infty \quad\Rightarrow\quad \prod\limits_{l=1}^\infty (1-a_k) = 0. \tag1\)

Note that the right conclusion is also true, if the a_l do not converge to 0, but are bounded by 1.

Now we simply apply all this with

\(a_k = \dfrac{1}{2k}\)

and get that the sum above is equal to 1/2.

For the records, we study the conclusion (1) in more detail. First of all for positive a_l between 0 and 1, we get

\(\sum\limits_{k=1}^\infty a_l = \infty \quad\Leftrightarrow\quad \prod\limits_{l=1}^\infty (1-a_l) = 0.\)

We can use the estimate

\(– 2h \le \log(1-h) \le -h \qquad\text{for } 0 \le h < \dfrac{1}{2}<h\)

for this. In case, infinitely many a(l) are larger that 1/2, the result is obvious.

In a similar way we can prove for positive a(l)

\(\sum\limits_{k=1}^\infty a_l = \infty \quad\Leftrightarrow\quad \prod\limits_{l=1}^\infty (1+a_l) < \infty.\)

If the a(l) negative and positive things get more involved.