자연계열 1-3번

[1.3] 급수와 수열의 극한 사이의 관계로부터 급수 $\displaystyle \sum_{n=1}^\infty\left(a_n-\frac{2n^2+2n-1}{n^2+n}\right)$가 수렴하면$\displaystyle \lim_{n\to\infty}\left(a_n-\frac{2n^2+2n-1}{n^2+n}\right)=0$이므로,$\displaystyle \begin{aligned}\lim_{n\to\infty}a_n&=\lim_{n\to\infty}\left(a_n-\frac{2n^2+2n-1}{n^2+n}+\frac{2n^2+2n-1}{n^2+n}\right)\\&=\lim_{n\to\infty}\left(a_n-\frac{2n^2+2n-1}{n^2+n}\right)+\lim_{n\to\infty}\frac{2n^2+2n-1}{n^2+n}\\&=0+2=2\end{aligned}$이다.$\displaystyle \sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)=1$이므로$\displaystyle \begin{aligned}\lim_{n\to\infty}\left(\sum_{k=1}^n(a_n-2)\right)&=\sum_{n=1}^\infty(a_n-2)\\&=\sum_{n=1}^\infty\left(a_n-\frac{2n^2+2n}{n^2+n}\right)\\&=\sum_{n=1}^\infty\left(a_n-\frac{2n^2+2n-1}{n^2+n}-\frac1{n^2+n}\right)\\&=\sum_{n=1}^\infty\left(a_n-\frac{2n^2+2n-1}{n^2+n}\right)-\sum_{n=1}^\infty\frac1{n^2+n}=5-1=4\end{aligned}$이고, 따라서$\displaystyle \lim_{n\to\infty}\left\{a_n+\sum_{k=1}^n(a_k-2)\right\}=\lim_{n\to\infty}a_n+\lim_{n\to\infty}\left(\sum_{k=1}^n(a_k-2)\right)=2+4=6$이다.