参考にしたサイト➔これだよ
解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \lim_{n \to \infty} \left(\dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \cdots + \dfrac{1}{2n} \right) & = & \displaystyle \lim_{n \to \infty} \displaystyle \sum_{k=1}^{n} \dfrac{1}{n + k} \\ & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^{n} \dfrac{1}{1 + \dfrac{k}{n}} \\ & = & \int_{0}^{1} \dfrac{1}{1 + x} dx \\ & = & \log2 \end{eqnarray} \)解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \lim_{n \to \infty} \left(\dfrac{n}{n^2 + 1} + \dfrac{n}{n^2 + 4} + \cdots + \dfrac{1}{2n} \right) & = & \displaystyle \lim_{n \to \infty} \displaystyle \sum_{k=1}^{n} \dfrac{n}{n^2 + k^2} \\ & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^{n} \dfrac{n}{n + \dfrac{k^2}{n}} \\ & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^{n} \dfrac{1}{1 + \left(\dfrac{k}{n} \right)^2} \\ & = & \int_{0}^{1} \dfrac{1}{1 + x^2} dx \\ & = & \dfrac{\pi}{4} \end{eqnarray} \)解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \left(\dfrac{1}{\sqrt{n^2 + 1}} + \dfrac{2}{\sqrt{n^2 + 4}} + \cdots + \dfrac{n - 1}{\sqrt{2n^2 - 2n + 1}} \right) & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^{n} \dfrac{n}{\sqrt{n^2 + k^2}} - \dfrac{1}{\sqrt{2}n} \\ & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^{n} \dfrac{1}{\sqrt{1 + \left( \dfrac{k}{n} \right)^2}} \\ & = & \int_{0}^{1} \dfrac{1}{\sqrt{1 + x^2}} dx \\ & = & \sqrt{2} - 1 \end{eqnarray} \)解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \lim_{n \to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n} ke^{\frac{k}{n}} & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \sum_{k=1}^{n} \dfrac{k}{n}e^{\frac{k}{n}} \\ & = & \int_{0}^{1} xe^x dx \\ & = & 1 \\ \end{eqnarray} \)解説：
\(\displaystyle \lim_{n \to \infty} \dfrac{\sqrt[n]{(n + 2)(n + 4) \cdots (3n)}}{n} \)を\( A \)とおくと
\( \hspace{35px} \begin{eqnarray} \displaystyle \lim_{n \to \infty} \log A & = & \displaystyle \lim_{n \to \infty} \log \sqrt[n]{\left (1+\dfrac{2}{n} \right) \left(1+\dfrac{4}{n} \right) \cdots \left(1+\dfrac{2n}{n} \right)} \\ & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \log \left(1+\dfrac{2}{n} \right) \left(1+\dfrac{4}{n} \right) \cdots \left(1+\dfrac{2n}{n} \right) \\ & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \left\{\log \left( 1 + \dfrac{2}{n} \right) + \log \left(1 + \dfrac{4}{n} \right) + \cdots + \log \left( 1 + \dfrac{2n}{n} \right)\right\} \\ & = & \displaystyle \lim_{n \to \infty} \dfrac{1}{n}\sum_{k=1}^{n}\log \left( 1 + 2\dfrac{k}{n} \right) \\ & = & \int_{0}^{1} \log(1 + 2x) dx \\ & = & \dfrac{3}{2}\log 3 - 1 \end{eqnarray} \)