# Thread: Using the alternating series test for convergence ?

1. ## Using the alternating series test for convergence ?

The alternating series test is given as:

$\sum^{\infty}_{r=1}(-1)^{r-1}b_r$ , where $b_n > 0$ , converges if $(b_n)$ is a monotone decreasing sequence with limit zero.

I have 3 problems,I am asked to investigate these for conditional or absolute convergence, or divergence.

a) $\sum^{\infty}_{r=1}(-1)^r(\sqrt{r(r+1)}-r)$

b) $\sum^{\infty}_{r=1}\frac{(-1)^r r^{3/2}}{r^2-1}$

c) $\sum^{\infty}_{r=1}\frac{(-1)^r r}{e^r}$

for problem a) :

I think i should transform (a) into the form $\sum^{\infty}_{r=1}(-1)^{r-1}(r-\sqrt{r(r+1)})$ and then prove that $(r-\sqrt{r(r+1)})$ is positive, and a monotone decreasing sequence with limit zero, to show that (a) converges. But how exactly can i show $\lim_{n\rightarrow \infty}(n-\sqrt{n(n+1)}) = 0$ ? Don't know how to find the limit in this form ! Also,to test for absolute convergence of the series, do i merely test the series but exclude the factor $(-1)^r$ ? i.e to test for absolute convergence do i test the convergence of:

a) $\sum^{\infty}_{r=1}(\sqrt{r(r+1)}-r)$ ?

for b) :

Here r cannot be 1! do i just check from r=2? Getting (b) into the form to use the alternating series test i get
$\sum^{\infty}_{r=1}\frac{(-1)^{r-1}r^{3/2}}{1-r^2}$ which leaves $b_n=\frac{n^{3/2}}{1-n^2}$ to be proven positive,and decreasing to limit zero. I thought about multiplying both numerator and denominator by n,to divide all terms by a dominant term $n^3$ ? to get $\frac{n^(5/2)}{n-n^3}\div n^3 = \frac{n^{-\frac{1}{2}}}{\frac{1}{n^2}-1}$
This seems to tend to zero as n tends to infinity. But now for absolute convergence, do i test $\sum^{\infty}_{r=1}\frac{r^{3/2}}{r^2-1}$ to see if this converges ?

2. For #a note that $\displaystyle \sqrt{r(r+1)}-r=\frac{1}{\sqrt{ 1+\frac{1}{r}}+1}$

3. It took me a while to see how this helps, but thanks !

4. Originally Posted by punkstart
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But now for absolute convergence, do i test $\sum^{\infty}_{r=1}\frac{r^{3/2}}{r^2-1}$ to see if this converges ?
For large $r$ the absolute value of the terms behave as $r^{-1/2}$ which means that the series cannot converge absolutely.

CB

5. Originally Posted by punkstart
c) $\sum^{\infty}_{r=1}\frac{(-1)^r r}{e^r}$
for all $x>0:\ e^x>\dfrac{x^3}{6}$

CB