## University Calculus: Early Transcendentals (3rd Edition)

Here, $\Sigma_{n=1}^{\infty} a_n=\Sigma_{n=1}^{\infty} (\dfrac{1}{n}-\dfrac{1}{n+1})$ and $s_n= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-.....+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}$ Thus, we have the nth partial sum: $(1-\dfrac{1}{n})$. Now, we see that $s_n \to 1$ as $n \to \infty$, so the given series will converge and the sum will be $1-\lim\limits_{n \to \infty}\dfrac{1}{n}=1$.