Proving convergence or divergence of a series.

[kilroymcb]

I'm asked to find out if this series is convergent, divergent or conditionally convergent.

Sum from 1 to infinity of ((-1)^(n+1))/4th root of n.
I used the ratio test and came out with L = 1... which is inconclusive. I thought about using the alternating series test, but since this isnt that section, I thought that might not be the solution they were looking for. I think this is conditionally convergent, the problem is, how to prove it?

Here is another one.
Sum f rom 1 to infinity of (1)/(2N)!. I did the ratio test for Abs value of 1/(2(n+1))! * (2n)!/1. This goes to (2n)!/(2(n+1))! which = 2n/2(n+1). lim as n goes to infinity of 2n/2n+2 = 1/1 + 2/2n. 1/1+0 = 1. This is inconclusive... but the book claims this is absolutely convergent. How?

Also, sum from 1 to infinity of ((-1)^n e^(1/n))/n^3. Ratio test,(( e^1/(n+1))/((n+1)^3)) * (n^3)/e^(1/n). Wouldnt you end up with e^1 in the numerator *n^3 all divided by (n+1)^3?

 

[pka]

Do you know the alternating series test?
Given \(\displaystyle a_n = \frac{1}{{\sqrt[4]{n}}}\) can you prove that \(\displaystyle a_n\) is a decreasing null sequence?

 

[kilroymcb]

Do you know the alternating series test?
Given \(\displaystyle a_n = \frac{1}{{\sqrt[4]{n}}}\) can you prove that \(\displaystyle a_n\) is a decreasing null sequence?

Yes, I know the alternating series test. However, this problem is given in the section dealing with Ratio and root tests. Considering that, I would think they would want me to use one of those tests to solve the problem... Unless you are saying that I need to do the alternating series test after I discover that L = 1?

Anyway, any sum of the form 1/a^n is decreasing. The limit as n goes to infinity of 1/a^n is 0, providing n is non-negative. By AST, that would be convergent.

 

[soroban]

Hello, kilroymcb!

In the second one, you didn't reduce correctly . . .

\(\displaystyle \L\sum_{n=1}^{\infty}\frac{1}{(2n)!}\)

\(\displaystyle \L\frac{a_{n+1}}{a_n} \:=\:\frac{1}{(2[n\,+\,1])!}\,\cdot\,\frac{(2n)!}{1} \;=\;\frac{(2n)!}{(2n+2)(2n+1)(2n)!} \;=\;\frac{1}{(2n+2)(2n+1)\)

Then: \(\displaystyle \L\:\lim_{n\to\infty}\left|\frac{1}{(2n+2)(2n+1)}\right| \;=\;0\)

\(\displaystyle \L\sum_{n=1}^{\infty}\frac{(-1)^n e^{\frac{1}{n}}}{n^3}\)

\(\displaystyle \L\frac{a_{n+1}}{a_n} \;=\;\frac{e^{\frac{1}{n+1}}}{(n+1)^3}\,\cdot\,\frac{n^3}{e^{\frac{1}{n}}} \;=\;\frac{n^3}{(n+1)^3}\cdot\frac{e^{\frac{1}{n}}} {e^{\frac{1}{n+1}}}\)

Note that: \(\displaystyle \L\:\frac{e^{\frac{1}{n}}}{e^{\frac{1}{n+1}}} \;=\;e^{(\frac{1}{n} -\frac{1}{n+1})} \;=\;e^{\frac{1}{n(n+1)}}\;\;\) . . . subtract exponents, remember?

So the ratio becomes: \(\displaystyle \L\:\left(\frac{n}{n+1}\right)^3 e^{\frac{1}{n(n+1)}}\)

Unfortunately, the limit goes to \(\displaystyle 1\) . . . inconclusive.

We must find another test . . .

 

[kilroymcb]

If we do part 1 of the AST on (((-1)^n)e^1/n))/n^3
Lim as n approaches infinity (e^1/n)/n^3 = 1/infinity which is 0. So, that is satisfied. If we do part two to try and find if it is decreasing...
(e^1/n+1)/(n+1)^3 < (e^1/n)/(n)^3. Therefore, it would converge by AST.

 

[skeeter]

for the last series where the ratio test failed to determine convergence/divergence, how about a direct comparison?

for n > 1 ...

\(\displaystyle \L \sum \frac{e^{\frac{1}{n}}}{n^3} < \sum \frac{e}{n^3} = e \cdot \sum \frac{1}{n^3}\)

the series converges absolutely.

 

[pka]

If we do part 1 of the AST on (((-1)^n)e^1/n))/n^3
Lim as n approaches infinity (e^1/n)/n^3 = 1/infinity which is 0. So, that is satisfied. If we do part two to try and find if it is decreasing...
(e^1/n+1)/(n+1)^3 < (e^1/n)/(n)^3. Therefore, it would converge by AST.

Yes that is the idea, because the root/ratio test fails.
If you have not yet studied absolute convergence.

 

[kilroymcb]

If we do part 1 of the AST on (((-1)^n)e^1/n))/n^3
Lim as n approaches infinity (e^1/n)/n^3 = 1/infinity which is 0. So, that is satisfied. If we do part two to try and find if it is decreasing...
(e^1/n+1)/(n+1)^3 < (e^1/n)/(n)^3. Therefore, it would converge by AST.

Yes that is the idea, because the root/ratio test fails.
If you have not yet studied absolute convergence.

In order to find out if it converges absolutely, I'd just have to redo the AST but with absolute value signs, correct?