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December 4th, 2017, 07:09 AM  #1 
Newbie Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0  Convergence/divergence: strange case
Everyone knows that $\sum \frac{1}{n^a}$ is divergent if $a\leq 1$ and is convergent if $a >1$. But, I don't remember who said that thing: if I have this sum $\sum \frac{1}{n^{b_n}}$ if $b_n \longrightarrow 1$ I can't say instantly if this sum is convergent or divergent. Is it true? Can you make me an example? Last edited by Berker; December 4th, 2017 at 07:11 AM. 
December 4th, 2017, 08:54 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra 
Well, the trivial case $b_n=1$ for all $n$ obviously doesn't converge. Nor does any case $b_n \le 1$ for all $n$. For convergence by the Root Test, we need \begin{align*}\displaystyle \lim_{n \to \infty} \leftn^{b_n}\right^{\frac1n} & < 1 \\ \lim_{n \to \infty} n^{\frac{b_n}{n}} & < 1 \\ \log \lim_{n \to \infty} n^{\frac{b_n}{n}} & < 0 \\ \lim_{n \to \infty} \log n^{\frac{b_n}{n}} & < 0 \\ \lim_{n \to \infty} b_n \frac{\log n}{n} & < 0 \\ \left(\lim_{n \to \infty} b_n\right) \left(\lim_{n \to \infty} \frac{\log n}{n}\right) & < 0 \end{align*} But the left hand limit is always equal to zero (when the $b_n$ converge). So the root test will never give convergence. That doesn't mean we can't get convergence though. My gut feeling is that divergence is guaranteed. Last edited by v8archie; December 4th, 2017 at 09:29 AM. 
December 4th, 2017, 09:26 AM  #3 
Newbie Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 
And for $b_n \longrightarrow 1^{+}$, (so $b_n>1$) is it possible that the sum converges?

December 4th, 2017, 10:50 AM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 227 Thanks: 122 Math Focus: Dynamical systems, analytic function theory, numerics 
This series can't converge for any choice of $b_n$. To see this, write $b_n = 1 + c_n$ so that now $c_n \to 0$. The sum you have written is equivalent to $\zeta(1+c_n)$ whenever $c_n >0$ is real where $\zeta$ is \[ \zeta(s) = \sum_{n=1}^\infty n^{s} \quad \text{Re}(s) > 1. \] Now, at $s = 1$ we have exactly the usual harmonic series implying the sum blows up. In fact, this is independent of the sum formula since $\zeta$ admits an analytic continuation which is analytic on the entire complex plane except at a simple pole precisely when $s= 1$. Thus, $s = 1$ is a simple pole requires $\lim_{s \to 1} \left\zeta(s) \right = \infty$. This can be seen directly from the residue theorem applied to our sequence, $c_n$ which gives \[ \lim_{n\to \infty} c_n \zeta(1+c_n) \to 1 \] and since $c_n \to 0$, we must have $\zeta(1+c_n) = \zeta(b_n) \to \infty$. 

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case, convercence or divergence, convergence or divergence, strange 
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