Answer
$\qquad \displaystyle \sum_{n=1}^{\infty}\frac{n^{2}}{n^{2}+1}$
Work Step by Step
From the first four terms, we observe the pattern
nth numerator = $n^{2}$
nth denominator = (numerator)+1 = $n^{2}+1$
So, the nth term is $\displaystyle \frac{n^{2}}{n^{2}+1}.$
The first term corresponds to n=1 (lower limit).
The sum has infinitely many terms (no upper limit ... we write $\infty$)
Summation notation:$\qquad \displaystyle \sum_{n=1}^{\infty}\frac{n^{2}}{n^{2}+1}$
