\section*{Problem 1 - 171231}

Given the sequence \(\left(a_{n}\right)\) by

\[
a_{n}=\frac{4 n}{4 n^{2}+121}
\]

for \(n=1,2,3, \ldots\).

Determine the upper limit and the lower limit of \(\left(a_{n}\right)\), if they exist.
}

1. Lower Bound:

\(\frac{4 n}{4 n^{2}+121}\) is, since \(n\) is a natural number, always positive. If \(n\) approaches infinity, we have:

\[
\lim _{n \rightarrow \infty} \frac{4 n}{4 n^{2}+121}=\lim _{n \rightarrow \infty} \frac{\frac{4}{n}}{4+\frac{121}{n^{2}}}=\frac{0}{0+4}=0
\]

\(\left(a_{n}\right)\) is thus a null sequence. Since all its terms are positive, 0 is the lower bound (which never occurs in the sequence).

2. Upper Bound:

Subtract \(a_{n+1}\) from \(a_{n}\):

\[
\begin{gathered}
a_{n}-a_{n+1}=\frac{4 n}{4 n^{2}+121}-\frac{4(n+1)}{4(n+1)^{2}+121}=\frac{4 n \cdot\left(4 n^{2}+8 n+125\right)-(4 n+4) \cdot\left(4 n^{2}+121\right)}{\left(4 n^{2}+121\right) \cdot\left(4(n+1)^{2}+121\right)}= \\
=\frac{16 n^{2}+16 n-484}{\left(4 n^{2}+121\right) \cdot\left((2 n+2)^{2}+121\right)}
\end{gathered}
\]

The denominator is always positive since \(n\) is a natural number. The numerator increases as \(n\) increases. The only valid root of the numerator lies between \(n=5\) and \(n=6\):

\[
\begin{array}{lc}
n=5: & 16 n^{2}+16 n-484=-40
\end{array}
\]

For all \(n \leq 5\), \(16 n^{2}+16 n-484\) and thus \(a_{n}-a_{n+1} < 0 \Leftrightarrow a_{n+1} > a_{n}\). For \(n > 5\), \(16 n^{2}+16 n-484\) and thus \(a_{n}-a_{n+1} > 0 \Leftrightarrow a_{n+1} < a_{n}\). \(\left(a_{n}\right)\) is therefore strictly decreasing up to \(a_{6}\).

Consequently, the sequence attains its maximum value at \(a_{6}\). \(a_{6}=\frac{24}{265}\) is the upper bound of the sequence.