Question

Finding convergence using ratio test:
Let {a_n} (n=1)^inf be defined by
a_n = [(1/(n^n) if n is odd) AND (1/((2n^(2n)) if n is even)]

The answer should be inconclusive.. To solve this, do I need to add anything or can I just complete the ratio test and figure out if it converges or not?

I was thinking of multiplying in (2n+1)! to the odd series and (2n)! to the even series. Am I supposed to do this?
Also, if one series turns out to diverge then is the entire function considered inconclusive?

Thanks!

Answer 1

So you're given \(\displaystyle\sum_{n=1}^\infty a_n\), where \(\left\{a_n\right\}\) is defined by
\[a_n=\begin{cases}\dfrac{1}{n^n}&\text{for odd }n\\
\dfrac{1}{(2n)^{2n}}&\text{for even }n\end{cases}\]
Is that correct? It's hard to tell what your parentheses are saying for the even \(n\).

Answer 2

yeah, @SithsAndGiggles that is what I meant to write.. i just realized my parentheses were pretty off my bad

Answer 3

Firstly, I'd split up the given series into two:
\[\begin{align*}\sum_{n=1}^\infty a_n&=\sum_{k}b_k+\sum_{k}c_k\\
&=\sum_{k=0}^\infty \frac{1}{(2k+1)^{2k+1}}+\sum_{k=1}^\infty \frac{1}{(2(2k))^{2(2k)}}\\
&=\sum_{k=0}^\infty \frac{1}{(2k+1)^{2k+1}}+\sum_{k=1}^\infty \frac{1}{(4k)^{4k}}
\end{align*}\]
If you didn't quite catch on to the notation, basically what I did was write an explicit series for the odd and even \(n\) and switched the index to \(k\). Notice that plugging in \(k=0\) for the first series gives you the first term of \(a_n\), and plugging in \(k=1\) for the second series gives you the second term of \(a_n\).

Now let's try the ratio test on one of the series. I'll try \(\sum b_k\) first:
\[\lim_{k\to\infty}\left|\frac{1}{(2(k+1)+1)^{2(k+1)+1}}\cdot\frac{(2k+1)^{2k+1}}{1}\right|\\
\lim_{k\to\infty}\frac{(2k+1)^{2k+1}}{(2k+3)^{2k+3}}\\
\lim_{k\to\infty}\frac{2k+1}{(2k+3)^3}\cdot\left(\frac{2k+1}{2k+3}\right)^{2k}\\
\lim_{k\to\infty}\frac{2k+1}{(2k+3)^3}\cdot\lim_{k\to\infty}\left(\frac{2k+1}{2k+3}\right)^{2k}\]
I'll leave the remaining limit computation up to you, but you should find that \(\sum b_k\) does indeed converge by the ratio test.

Now for \(\sum c_k\):
\[\lim_{k\to\infty}\left|\frac{1}{(4(k+1))^{4(k+1)}}\cdot\frac{(4k)^{4k}}{1}\right|\\
\lim_{k\to\infty}\frac{(4k)^{4k}}{(4k+4)^{4k+4}}\\
\lim_{k\to\infty}\frac{1}{(4k+4)^4}\cdot\left(\frac{4k}{4k+4}\right)^{4k}\\
~~~~~~~~~~~~~~~~~~~~~\vdots\]
Similarly, you'll find that \(\sum c_k\) converges by the ratio test. So, since \(\sum a_n\) is the sum of two convergent series, \(\sum a_n\) must also converge.

You say the ratio test is inconclusive, though, which leads me to believe that the given series was interpreted incorrectly on my part...