## Precalculus (10th Edition)

We know that $e^{\ln{n}}=n$, In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(e^{\ln{n+1}})-e^{\ln{n}}=(n+1)-(n)=1$, thus it is an arithmetic sequence. $a_1=1$ $a_2=2$ $a_3=3$ $a_4=4$
We know that $e^{\ln{n}}=n$, In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(e^{\ln{n+1}})-e^{\ln{n}}=(n+1)-(n)=1$, thus it is an arithmetic sequence. $a_1=1$ $a_2=2$ $a_3=3$ $a_4=4$