11.1. Find the largest term of the sequence a) $a_{n}=\frac{n}{n^{2}+2020}$, б) $a_{n}=\frac{2020^{n}}{n!}$ (where $n!=1 \cdot 2 \cdot 3 \cdots n)$
Answer. a) $a_{45}=\frac{45}{4045}$, b) $a_{2019}=a_{2020}=\frac{2020^{2019}}{2019!}$. Solution. a) Let's investigate the inequality $a_{n}<a_{n+1}$, i.e., $a_{n}=\frac{n}{n^{2}+2020}<\frac{n+1}{(n+1)^{2}+2020}=a_{n+1}$. Transforming it, we get $n^{2}+n<2020$. This inequality holds, as it is easy to verify, for $n=44$ and does not hold for $n=45$. Due to the monotonically increasing function $f(n)=n^{2}+n$ (for positive $n$), $a_{45}$ is the largest term of the sequence. Remark. Another way to solve this is by studying the formula for the general term using the derivative. For the found maximum point $x$, the value of the function must be compared at the two nearest integer points to $x$ on the left and right. b) Similarly, comparing the expressions for $a_{n}$ and $a_{n+1}$ and reducing the inequality by $\frac{2020^{n}}{n!}$, we get $n+1<2020$. Thus, $a_{n}$ increases for $n<2019$; the values of $a_{2019}$ and $a_{2020}$ coincide, and for $\quad n \geq 2020$ the sequence $a_{n}$ decreases.