To what does the following sequence converge?

$$
a_{n}=\frac{\sum_{j=0}^{\infty}\binom{n}{2 j} \cdot 2^{j}}{\sum_{j=0}^{\infty}\binom{n}{2 j+1} \cdot 2^{j}}
$$
The terms of the sequence in the task are obtained when we generate rational numbers close to $\sqrt{2}$ according to the following reasoning. We start from the number $q=1-\sqrt{2}$, and we say that its $n$-th power

$$
q^{n}=(1-\sqrt{2})^{n} \sim 0
$$

Expanding the power using the binomial theorem, every second term contains $\sqrt{2}$:

$$
q^{n}=(1-\sqrt{2})^{n}=\sum_{j=0}^{n}\binom{n}{j}(-\sqrt{2})^{j}=\sum_{k=0}^{\infty}\binom{n}{2 k} \cdot 2^{k}-\sqrt{2} \cdot \sum_{k=0}^{\infty}\binom{n}{2 k+1} \cdot 2^{k}
$$

(For simplicity, we wrote $\infty$ as the upper limit of the summation; indeed, the terms we arbitrarily included in the sum are equal to 0—since $\binom{n}{j}$ is defined to be 0 if $j>n$.) Substituting the above expression for $q^{n}$ into the approximate equality $q^{n} \sim 0$, and solving for the value of $\sqrt{2}$, we obtain the approximate equality $\sqrt{2} \sim a_{n}$. More precisely, it is true based on the above that

$$
a_{n}-\sqrt{2}=\frac{q^{n}}{\sum_{k=0}^{\infty}\binom{n}{2 k+1} \cdot 2^{k}}
$$

Here, the value of the denominator is at least 1, since the denominator is a sum of non-negative terms, and its first term is $n \geq 1$. The numerator, however, tends to 0 as $n$ tends to infinity, since $|q|<1$. Therefore, $a_{n}-\sqrt{2}$ tends to 0 as $n$ tends to infinity, which means that the limit of the sequence $a_{n}$ is $\sqrt{2}$.