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December 4th, 2017, 07:09 AM   #1
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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.
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December 4th, 2017, 08:54 AM   #2
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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} \left|n^{-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.
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December 4th, 2017, 09:26 AM   #3
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And for $b_n \longrightarrow 1^{+}$, (so $b_n>1$) is it possible that the sum converges?
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December 4th, 2017, 10:50 AM   #4
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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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