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Can anybody help me with the following problem?

Let $p>0$ and consider the series: $$\sum_{n=2}^\infty \frac{1}{n\log_2^p n}.$$ Find the values of $p$ for which the series is convergent.

user405381
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The Cauchy Condensation Test states that $\displaystyle\sum_{n=1}^{\infty} f(n)$ converges if and only if $\displaystyle\sum_{n=1}^{\infty} 2^n f(2^n) $ converges provided $f(n)$ is non-negative and non-increasing.

Now this is a good test to use here, since we have a logarithm of base 2. In this case, let's take $f(n) = \frac{1}{n \log_2^p n}$. This clearly satisfies the conditions of the test because the lower limit is 2.

Now let's find $2^n f(2^n)$: $$2^n f(2^n) = 2^n \left(\frac{1}{2^n \log_2^p (2^n)} \right) = \frac{1}{n^p}$$

So applying the test, we have:

$\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \log_2^p n}$ converges if and only if $ \displaystyle\sum_{n=2}^{\infty} \frac{1}{n^p}$ converges.

This is a standard series. It converges if $p>1$ and diverges if $p \leq 1$.

PhysicsMathsLove
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  • Thank you! I am however a little stuck on showing that the sequence is decreasing. I know I have to show that $a_{n+1}\leq a_n$, but I have trouble translating it into a statement I know to be true. Any suggestions? – user405381 May 19 '17 at 20:08
  • @user405381 You could show that, or alternatively note that $n\log_2^pn$ is a strictly increasing sequence for $n>2$. $\frac{1}{\text{increasing sequence}}$ is decreasing. But for your method: $\frac{a_{n+1}}{a_n}$ $ = \frac{n}{n+1}$ $\left( \frac{\log_2 (n)}{\log_2 (n+1} \right)^p$, which is clearly less than or equal to 1. – PhysicsMathsLove May 19 '17 at 20:51
  • But for both of those arguments you use that $\log_2 (n)$ is increasing. How do I show that? Is it valid to say that since $\log(n)$ is increasing so is $\log_2(n)$? – user405381 May 19 '17 at 21:10
  • Well, it is a property of the $\log$ function. $y= \log_2(x) \Rightarrow \frac{dy}{dx} = \frac{1}{x \log(2)}$ and for $x>0$, this is increasing. If the derivative is too much machinery, then I can only think of the hand-waivey argument: $y = \log_2(x) \Leftrightarrow x = 2^y$ from which it clearly follows that $x$ increases with $y$. – PhysicsMathsLove May 19 '17 at 23:02