# Alternating Series

I have an alternating series problem, and I am looking to find convergence, or divergence. The square roots are making this difficult for me to solve.

I began by moving the $(-1)^{n+1}$ outside of the fraction, and performed the Alternating Series Test. When I try to complete the test, I'm having problems due to the square root in the numerator and denominator. I would appreciate any help. $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} (3 \sqrt{n+1})}{\sqrt{n}+1}$$

Here is an image referring to the summation with $(-1)^{n+1}$ moved out... $$\sum_{n=1} \frac{3\sqrt{n+1}}{\sqrt{n}+1}$$

• Notice what the terms of the series converge to as $n\to\infty$ and think about the Divergence Test. – user142299 Apr 28 '14 at 1:49

Here's a trick which helps for a lot of problems similar to this: divide the numerator and denominator by a convenient power of $n$. In this case, we divide by $\sqrt n$ to obtain:
$$\sum \frac{3\sqrt{n+1}}{\sqrt n + 1} = \sum \frac{3\sqrt{n+1}/\sqrt{n}}{1+1/\sqrt{n}} = \sum\frac{3\sqrt{(n+1)/(n)}}{1+1/\sqrt{n}} = \sum\frac{3\sqrt{1+1/n}}{1+1/\sqrt{n}}.$$
In the limit as $n \to \infty$, notice that $1/n$ and $1/\sqrt n$ go to zero, so we conclude that the terms converge to $\frac{3 \cdot 1}{1} = 3$. What does this tell you about the convergence of the series?
Hint: Let $a_n = \frac{3 \sqrt{n+1}}{\sqrt{n}+1}$. Since $f(x,y) = \sqrt{x^2+y^2}$ is a metric, it must satisfy the triangle inequality. Hence, $\sqrt{n+1} \leq \sqrt{n} + 1$, Hence
$$a_n \leq 3$$