Determine if the series converges absolutely, converges, or diverges.
-$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \left( \sqrt { n ^ { 2 } + 1 } - n \right)$$
A) converges absolutely
B) diverges
C) converges conditionally

Question

Quizplus · Accepted Answer

To show that the series converges conditionally, we need to show that it converges but does not converge absolutely.

First, let's examine if the series converges. We can use the alternating series test, since the series is alternating and the absolute value of each term approaches zero as n approaches infinity. Let
$$a_n = \left( \sqrt { n ^ { 2 } + 1 } - n \right)$$
Then we have
$$\lim_{n\to\infty} a_n = \lim_{n\to\infty}\left( \sqrt { n ^ { 2 } + 1 } - n \right) = \lim_{n\to\infty}\left(\frac{1}{\sqrt { n ^ { 2 } + 1 } + n}\right) = 0$$
Since $a_n$ is decreasing and approaches zero, the series converges by the alternating series test.

Next, let's examine if the series converges absolutely. We have
$$|a_n| = \left| (-1)^n(\sqrt { n ^ { 2 } + 1 } - n) \right| = \sqrt { n ^ { 2 } + 1 } - n \geq \sqrt{n^2} - n = 0$$
Therefore, $\sum_{n=1}^\infty |a_n|$ diverges by the comparison test with the harmonic series.

Since the series converges but does not converge absolutely, it converges conditionally.

Determine If the Series Converges Absolutely, Converges, or Diverges ∑n=1∞(−1)n(n2+1−n)\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \left( \sqrt { n ^ { 2 } + 1 } - n \right)n=1∑∞​(−1)n(n2+1​−n)

Determine If the Series Converges Absolutely, Converges, or Diverges $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \left( \sqrt { n ^ { 2 } + 1 } - n \right)$