Answer
$\displaystyle\lim_{n \to \infty}\sum_{n=1}^{\infty}\left(\frac{i^{4}}{n^5}+\frac{i}{n^2}\right)=\int_0^1(x^4+x)dx=\frac{7}{10}$
Work Step by Step
$$\begin{align*}
\displaystyle\lim_{n \to \infty}\sum_{n=1}^{\infty}\left(\frac{i^{4}}{n^5}+\frac{i}{n^2}\right)&=\lim_{n \to \infty}\sum_{n=1}^{\infty}\left(\frac{i^{4}}{n^4}+\frac{i}{n}\right)\frac{1}{n}\\
&=\lim_{n \to \infty}\sum_{n=1}^{\infty}\left[\left(\frac{i}{n}\right)^4+\frac{i}{n}\right]\frac{1}{n}\\
&=\lim_{n \to \infty}\sum_{n=1}^{\infty}\left[\left(0+\frac{i}{n}\right)^4+\left(0+\frac{i}{n}\right)\right]\frac{1-0}{n}\\
&=\lim_{n \to \infty}\sum_{n=1}^{\infty}(x_i^4+x_i)\frac{1-0}{n}\\
&=\int_0^1(x^4+x)dx\\
&=\left[\frac{x^5}{5}+\frac{x^2}{2}\right]_0^1\\
&=\frac{7}{10}
\end{align*}$$
