Determine convergence or divergence of the alternating series.
-$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { n + \sqrt { n } } { n ^ { 2 } + 1 }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

We can use the alternating series test to show convergence. Let $a_n=\frac{n+\sqrt{n}}{n^2+1}$.

First, note that $a_n>0$ for all $n\geq 1$.

Next, we must show that $a_n$ is decreasing as $n$ increases. We have $a_n-a_{n+1}=\frac{n+\sqrt{n}}{n^2+1}-\frac{n+1+\sqrt{n+1}}{(n+1)^2+1}$. Simplifying, we get $$\begin{aligned} a_n-a_{n+1}&=\frac{(n+\sqrt{n})[(n+1)^2+1]-(n+1+\sqrt{n+1})(n^2+1)}{(n^2+1)[(n+1)^2+1]} \ &=\frac{n^3-n\sqrt{n}+n+\sqrt{n}-n^3-n^2\sqrt{n+1}+\sqrt{n+1}+n^2+2n\sqrt{n+1}+n+1}{(n^2+1)[(n+1)^2+1]} \ &=\frac{2n\sqrt{n+1}-n^2\sqrt{n+1}-n\sqrt{n}+\sqrt{n}-1}{(n^2+1)[(n+1)^2+1]}. \end{aligned}$$

We want to show that the numerator is positive for all $n\geq 1$. Note that $2n\sqrt{n+1}>n^2\sqrt{n+1}$ for $n\geq 1$, so the first two terms are nonnegative. Furthermore, $\sqrt{n}>1$ for $n\geq 2$, so $\sqrt{n}-1\geq 0$ for $n\geq 2$. It follows that $a_n-a_{n+1}>0$ for $n\geq 2$.

Finally, since $a_n$ is decreasing and $\lim_{n	o\infty}a_n=0$, the alternating series test guarantees that the series converges.

Determine Convergence or Divergence of the Alternating Series ∑n=1∞(−1)n+1n+nn2+1\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { n + \sqrt { n } } { n ^ { 2 } + 1 }n=1∑∞​(−1)n+1n2+1n+n​​

Determine Convergence or Divergence of the Alternating Series $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { n + \sqrt { n } } { n ^ { 2 } + 1 }$