$\mathrm{Az}$

$$
a_{n}=\frac{(n+3)^{2}+3}{n(n+1)(n+2)} \cdot \frac{1}{2^{n+1}}
$$

is a sequence defined by its general term. We form the sequence

$$
b_{n}=\sum_{k=1}^{n} a_{k}
$$

Determine the limit of the sequence $b_{n}$ as $n \rightarrow+\infty$.
I. solution. For $n=1,2,3$, respectively,

$$
\begin{array}{ll}
b_{1}=a_{1}=\frac{19}{24} & \left(=1-\frac{5}{24}=1-\frac{5}{3 \cdot 2 \cdot 2^{2}}\right) \\
b_{2}=b_{1}+a_{2}=\frac{15}{16} & \left(=1-\frac{1}{16}=1-\frac{6}{3 \cdot 4 \cdot 2^{3}}\right) \\
b_{3}=b_{2}+a_{3}=\frac{313}{320} & \left(=1-\frac{7}{320}=1-\frac{7}{4 \cdot 5 \cdot 2^{4}}\right)
\end{array}
$$

We see that the sequence $b_{n}$ is monotonically increasing - since $a_{n}>0$ - and $b_{3}$ approaches 1 from below. Based on this, we wrote the values of $b_{1}, b_{2}, b_{3}$ in the form $1-h$, where the value of the missing $h$ is $1 / 4,8,1 / 16,1 / 45,7$, decreasing from term to term, and $a_{4}>0$.

We can generalize this. For $n \geq 1$, we have

$$
2^{n+1} a_{n}=\frac{(n+3)^{2}+3}{n(n+1)(n+2)}
$$

which can be rewritten as

$$
2^{n+1} a_{n}=\frac{n^{2}+6(n+2)}{n(n+1)(n+2)}=\frac{n^{2}}{n(n+1)(n+2)}+\frac{6(n+2)}{n(n+1)(n+2)}
$$

This can be further simplified to

$$
2^{n+1} a_{n}=\frac{n}{(n+1)(n+2)}+\frac{6}{n(n+1)}=n\left(\frac{1}{n+1}-\frac{1}{n+2}\right)+6\left(\frac{1}{n}-\frac{1}{1+n}\right)
$$

Thus,

$$
2^{n+1} a_{n}=\left(1-\frac{1}{n+1}\right)-\left(1-\frac{2}{n+2}\right)+\frac{6}{n}-\frac{6}{n+1}=\frac{6}{n}-\frac{7}{n+1}+\frac{2}{n+2}
$$

Therefore,

$$
a_{n}=\frac{3}{n \cdot 2^{n}}-\frac{7}{(n+1) \cdot 2^{n+1}}+\frac{1}{(n+2) \cdot 2^{n}}
$$

This means that the original denominator of $a_{n}$, which includes the factors $n, n+1, n+2$, appears in only one term of the three-term sum.

The sum $b_{n}=a_{1}+a_{2}+\ldots+a_{n}$, when expanded according to (2), results in fractions whose denominators are of the form $k \cdot 2^{k}$. Generally, a fraction with a denominator of $k \cdot 2^{k}$ arises from the decomposition of three terms of the sequence $a_{i}$, specifically:

$$
\begin{aligned}
a_{k-2} & =\frac{3}{(k-2) \cdot 2^{k-2}}-\frac{7}{(k-1) \cdot 2^{k-1}}+\frac{4}{k \cdot 2^{k}}, \\
a_{k-1} & =\frac{3}{(k-1) \cdot 2^{k-1}}-\frac{7}{k \cdot 2^{k}}+\frac{4}{(k+1) \cdot 2^{k+1}}, \\
a_{k} & =\frac{3}{k \cdot 2^{k}}-\frac{7}{(k+1) \cdot 2^{k+1}}+\frac{4}{(k+2) \cdot 2^{k+2}}
\end{aligned}
$$

Since the coefficient of the $\frac{1}{k \cdot 2^{k}}$ term in the sum $a_{k-2}+a_{k-1}+a_{k}$ is $(4-7+3)=0$, the sum of fractions with a denominator of $k \cdot 2^{k}$ is 0 in the entire sum $a_{1}+a_{2}+\ldots+a_{n}$. This is true for all natural numbers $k$ for which $a_{k-2}$, $a_{k-1}$, and $a_{k}$ actually appear in the sum, i.e., for which $3 \leq k \leq n$. In addition to these, the sum that produces $b_{n}$ includes fractions with denominators of $k \cdot 2^{k}$ for $k=1,2$ and $k=n+1, n+2$, so $b_{n}$ is equal to the sum of these:

$$
\begin{gathered}
b_{n}=\frac{3}{1 \cdot 2}-\frac{7}{2 \cdot 2^{2}}+\frac{3}{2 \cdot 2^{2}}+\frac{4}{(n+1) \cdot 2^{n+1}}-\frac{7}{(n+1) \cdot 2^{n+1}}+\frac{4}{(n+2) \cdot 2^{n+2}}= \\
=1-\frac{3}{(n+1) \cdot 2^{n+1}}+\frac{4}{(n+2) \cdot 2^{n+2}}
\end{gathered}
$$

(For $k=1$, only $a_{1}$, for $k=2$, $a_{1}$ and $a_{2}$, for $k=n+1$, $a_{n-1}$ and $a_{n}$, and for $k=n+2$, only $a_{n}$ decompositions had to be considered in writing the above sum.)

From this, $\lim b_{n}=1$, because the second and third terms have a limit of 0.

Pallagi Dezsó (Budapest, Berzsenyi D. Gymnasium, III. grade) Kovács István (Budapest, I. István Gymnasium, IV. grade)

II. solution. Based on the identity $(n+3)^{2}+3=n^{2}+6(n+2)$, the term $a_{n}$ can be rewritten as:

$$
\begin{gathered}
2^{n+1} a_{n}=\frac{n^{2}}{n(n+1)(n+2)}+\frac{6(n+2)}{n(n+1)(n+2)}= \\
=\frac{n}{(n+1)(n+2)}+\frac{6}{n(n+1)}=n\left(\frac{1}{n+1}-\frac{1}{n+2}\right)+6\left(\frac{1}{n}-\frac{1}{1+n}\right)= \\
=\left(1-\frac{1}{n+1}\right)-\left(1-\frac{2}{n+2}\right)+\frac{6}{n}-\frac{6}{n+1}=\frac{6}{n}-\frac{7}{n+1}+\frac{2}{n+2}
\end{gathered}
$$

Thus,

$$
a_{n}=\frac{3}{n \cdot 2^{n}}-\frac{7}{(n+1) \cdot 2^{n+1}}+\frac{1}{(n+2) \cdot 2^{n}}
$$

This means that the original denominator of $a_{n}$, which includes the factors $n, n+1, n+2$, appears in only one term of the three-term sum.

The sum $b_{n}=a_{1}+a_{2}+\ldots+a_{n}$, when expanded according to (2), results in fractions whose denominators are of the form $k \cdot 2^{k}$. Generally, a fraction with a denominator of $k \cdot 2^{k}$ arises from the decomposition of three terms of the sequence $a_{i}$, specifically:

$$
\begin{aligned}
a_{k-2} & =\frac{3}{(k-2) \cdot 2^{k-2}}-\frac{7}{(k-1) \cdot 2^{k-1}}+\frac{4}{k \cdot 2^{k}}, \\
a_{k-1} & =\frac{3}{(k-1) \cdot 2^{k-1}}-\frac{7}{k \cdot 2^{k}}+\frac{4}{(k+1) \cdot 2^{k+1}}, \\
a_{k} & =\frac{3}{k \cdot 2^{k}}-\frac{7}{(k+1) \cdot 2^{k+1}}+\frac{4}{(k+2) \cdot 2^{k+2}}
\end{aligned}
$$

Since the coefficient of the $\frac{1}{k \cdot 2^{k}}$ term in the sum $a_{k-2}+a_{k-1}+a_{k}$ is $(4-7+3)=0$, the sum of fractions with a denominator of $k \cdot 2^{k}$ is 0 in the entire sum $a_{1}+a_{2}+\ldots+a_{n}$. This is true for all natural numbers $k$ for which $a_{k-2}$, $a_{k-1}$, and $a_{k}$ actually appear in the sum, i.e., for which $3 \leq k \leq n$. In addition to these, the sum that produces $b_{n}$ includes fractions with denominators of $k \cdot 2^{k}$ for $k=1,2$ and $k=n+1, n+2$, so $b_{n}$ is equal to the sum of these:

$$
\begin{gathered}
b_{n}=\frac{3}{1 \cdot 2}-\frac{7}{2 \cdot 2^{2}}+\frac{3}{2 \cdot 2^{2}}+\frac{4}{(n+1) \cdot 2^{n+1}}-\frac{7}{(n+1) \cdot 2^{n+1}}+\frac{4}{(n+2) \cdot 2^{n+2}}= \\
=1-\frac{3}{(n+1) \cdot 2^{n+1}}+\frac{4}{(n+2) \cdot 2^{n+2}}
\end{gathered}
$$

(For $k=1$, only $a_{1}$, for $k=2$, $a_{1}$ and $a_{2}$, for $k=n+1$, $a_{n-1}$ and $a_{n}$, and for $k=n+2$, only $a_{n}$ decompositions had to be considered in writing the above sum.)

From this, $\lim b_{n}=1$, because the second and third terms have a limit of 0.