Let a_1 = 1 and, for n 1, [a_n+1 = (a_n)/(2) + (2)/((n+1)^2) - (1)/(n^2).] Then -_n=1^(a_n…

Question

Let $a_1 = 1$ and, for $n \ge 1$,
$$a_{n+1} = \frac{a_n}{2} + \frac{2}{(n+1)^2} - \frac{1}{n^2}.$$
Then $-\displaystyle\sum_{n=1}^{\infty}\left(a_n - \frac{2}{n^2}\right)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

Let $b_n = a_n - \dfrac{2}{n^2}$. From the recurrence,
$$a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2}\left(a_n - \frac{2}{n^2}\right) \implies b_{n+1} = \frac{1}{2}b_n,$$
so $\{b_n\}$ is a G.P. with common ratio $\tfrac12$ and first term $b_1 = a_1 - 2 = -1$.

$$\sum_{n=1}^{\infty} b_n = \frac{b_1}{1 - \tfrac12} = \frac{-1}{1/2} = -2.$$

Hence $-\displaystyle\sum_{n=1}^{\infty}\left(a_n - \frac{2}{n^2}\right) = -(-2) = 2$.

Solutions · 1