# Thread: Test the series for convergence or divergence.

1. ## Test the series for convergence or divergence.

∑ (-1)^n n / √(n³ + 2)
n=1

2. Originally Posted by fardeen_gen

∑ (-1)^n n / √(n³ + 2)
n=1
By Leibnitz's Criterion this converges because if we define $\displaystyle a_n=\frac{n}{\sqrt{n^3+2}}$ then $\displaystyle a_{n+1}\leqslant{a_n}$ and $\displaystyle \lim_{n\to\infty}a_n=0$

3. Hello, fardeen_gen!

Test for convergence: .$\displaystyle \sum^{\infty}_{n=1}(-1)^n\,\frac{n}{\sqrt{n^3+2}}$

The absolute value of the $\displaystyle n^{th}$ term is: .$\displaystyle |a_n| \:=\:\frac{n}{\sqrt{n^3+2}}$

$\displaystyle \text{Divide top and bottom by }\sqrt{n^3}\!:\;\;\frac{\dfrac{n}{\sqrt{n^3}}} {\dfrac{\sqrt{n^3+2}}{\sqrt{n^3}}} \;=\;\frac{\dfrac{n}{n^{\frac{3}{2}}}} {\sqrt{\dfrac{n^3+2}{n^3}}}$ .$\displaystyle = \;\frac{\dfrac{1}{n^{\frac{1}{2}}}} {\sqrt{1 + \dfrac{2}{n^3}}}$

Then: .$\displaystyle \lim_{n\to\infty}|a_n| \;=\;\lim_{n\to\infty}\left[ \frac{\dfrac{1}{n^{\frac{1}{2}}}} {\sqrt{1 + \dfrac{2}{n^3}}}\right] \;=\;\frac{0}{1} \;=\;0$

We have an alternating series whose $\displaystyle n^{th}$ term goes to 0.

. . Therefore, the series converges.