Answer
See explanation.
Work Step by Step
(a) $\lim _{n \rightarrow+\infty} \frac{1}{n} \sum_{k=1}^{n} 1=\lim _{n \rightarrow+\infty} 1=\frac{n}{n}=1$
(b) $\lim _{n \rightarrow+\infty} \frac{1}{n^{2}} \sum_{k=1}^{n} k=\lim _{n \rightarrow+\infty} \frac{(1+n)n}{2 n^{2}}=\frac{1}{2}$
(c) $\lim _{n \rightarrow+\infty} \frac{1}{n^{3}} \sum_{k=1}^{n} k^{2}=\lim _{n \rightarrow+\infty} \frac{(1+n)(1+2 n)n}{6 n^{3}}=\frac{2}{6}=\frac{1}{3}$
(d) $\lim _{n \rightarrow+\infty} \frac{1}{n^{4}} \sum_{k=1}^{n} k^{3}=\lim _{n \rightarrow+\infty} \frac{(1+n)^{2}n^{2}}{4 n^{4}}=\frac{1}{4}$
