# Thread: absolute or conditional convergence

1. ## absolute or conditional convergence

Does this series converge absolutely or conditionally:
$\displaystyle \sum _{n=1}^\infty (-1)^n \frac{n^3+5^n}{(n+2)!}$

Problem (1): how do I even apply the alternating test when the demoninator is $\displaystyle (n+2)!$?
Problem (2): lets assume we figured out problem (1) which test would I use afterward to see if it converges absolutely?

Does this series converge absolutely or conditionally:
$\displaystyle \sum _{n=1}^\infty (-1)^n \left(\frac{1}{\ln n}\right)^4$

Soultion (1): Ok this time, the series does converge by alternating series test...
Problem (2): Which test do I use to see if it converges absolutely? I tried using comparision test: $\displaystyle \ln n < n \Longrightarrow \frac{1}{\ln n}>\frac{1}{n} \Longrightarrow \frac{1}{(\ln n)^4}>\frac{1}{n^4}$
$\displaystyle \frac{1}{n^4}$ converges (p-series with p>1) but this has no conclusion on whether $\displaystyle \left(\frac{1}{\ln n}\right)^4$ converges or not.

How do I solve these question?!? Any help appreciated!

2. Hello
Originally Posted by polymerase
Does this series converge absolutely or conditionally:
$\displaystyle \sum _{n=1}^\infty (-1)^n \frac{n^3+5^n}{(n+2)!}$

Problem (1): how do I even apply the alternating test when the demoninator is $\displaystyle (n+2)!$?
Here, I'd split the series into $\displaystyle \sum _{n=1}^\infty (-1)^n \frac{n^3}{(n+2)!}+\sum _{n=1}^\infty (-1)^n \frac{5^n}{(n+2)!}$ and show that each one converges absolutely/alternately.
To apply the alternating test you have to show, for example for the first series, that $\displaystyle (U_n)$ decreases (with $\displaystyle U_n=\frac{n^3}{(n+2)!}$) :
$\displaystyle \frac{U_{n+1}}{U_n}=\frac{ \frac{(n+1)^3}{(n+3)!}}{\frac{n^3}{(n+2)!}} =\frac{\left(1+\frac{1}{n}\right)^3}{n+3}< 1$ for $\displaystyle n$ sufficiently large.
Given that $\displaystyle 0\leq \frac{n^3}{(n+2)!}\leq \frac{1}{(n+2)(n+1)(n-3)!}$
The RHS tends to 0 as $\displaystyle n$ approaches $\displaystyle \infty$, the squeeze theorem tells us $\displaystyle (U_n)$ does the same and you can conclude.

Note : this series converges absolutely so I wrote this because you asked it but showing directly absolute convergence is quicker.
Problem (2): lets assume we figured out problem (1) which test would I use afterward to see if it converges absolutely?
$\displaystyle 0\leq \frac{n^3}{(n+2)!}\leq \frac{1}{(n+2)(n+1)(n-3)!}\leq\frac{1}{(n-3)!}$ for $\displaystyle n$ sufficiently large and $\displaystyle \sum (-1)^nU_n$ converges absolutely by comparison to the exponential series... (the same idea applies for the other part of the series, compare it to an exponential series)

Does this series converge absolutely or conditionally:
$\displaystyle \sum _{n=1}^\infty (-1)^n \left(\frac{1}{\ln n}\right)^4$

Problem (2): Which test do I use to see if it converges absolutely? I tried using comparison test: $\displaystyle \ln n < n \Longrightarrow \frac{1}{\ln n}>\frac{1}{n} \Longrightarrow \frac{1}{(\ln n)^4}>\frac{1}{n^4}$
$\displaystyle \frac{1}{n^4}$ converges (p-series with p>1) but this has no conclusion on whether $\displaystyle \left(\frac{1}{\ln n}\right)^4$ converges or not.
Shoudn't the series start with $\displaystyle k=2$ and not 1 ?
$\displaystyle t \mapsto \frac{1}{\ln^4t}$ decreases for $\displaystyle t>1$ so $\displaystyle \frac{1}{\ln^4n}\geq\int_n^{n+1}\frac{1}{\ln^4t}\, \mathrm{d}t$

We get $\displaystyle \sum_{k=2}^n\frac{1}{\ln^4k}\geq\int_2^{n+1}\frac{ 1}{\ln^4t} \,\mathrm{d}t$

Substituting $\displaystyle u=\ln t$, one can show that the integral tends to $\displaystyle \infty$ when n approaches $\displaystyle \infty$... hence, the series doesn't converge absolutely.