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I Need to show that

$$ \sum \frac{1}{(\log n)^a}$$

Diverges for all positive values of a.

My idea is to use the comparasion test, since the ratio test and rpot test are inconclusive. I want tp show that

$$ n^a > ln(n)$$ for values of n bigger than one $n_o$. This looks like a good approach, but i coulsnt find the $n_o$

Thanks in advance

Giiovanna
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4 Answers4

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hint: compare with Harmonic series, which diverges

EDIT: consider the function $\frac{\log^k n}{n}$, set $\log n =t, \ n = e^t$. We need to show $a_n = \frac{e^n}{n^k}$ is an increasing function for some $n(k)$. Consider the ratio $\frac{a_{n+1}}{a_n}$. After a bit of algebra you can show it is larger than $1$ for $n >\frac{e^{-\frac{1}{k}}}{1-e^{-\frac{1}{k}}}$.

Hence, $e^n >n^k \ \forall \ n> \frac{e^{-\frac{1}{k}}}{1-e^{-\frac{1}{k}}} \Rightarrow n >\log^k n \Rightarrow \frac{1}{n}<\frac{1}{\log^k n}$

Alex
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You don't need to find the exact $n_0$ if you use l'Hopital's rule. Suppose you want to take $a = 1$. Then by l'Hopital $$\lim_{n \rightarrow \infty} {\ln n \over n} = \lim_{n \rightarrow \infty } {{1 \over n} \over 1} = 0$$ So there will be some $n_0$ for which ${\displaystyle {\ln n \over n} < 1}$ for all $n > n_0$, which is all you need here.

Zarrax
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  • thanks for your answer. I was thinking to get the serie$ \sum \frac{1}{n^a}$, right? We will have that $\lim (\frac{ln(n)}{n})^a $ will be zero for all the values of a>0. But since $\sum \frac{1}{n^a}$ converges for a>1, doesnt that implies the convergence of the serie for a>1? – Giiovanna Aug 26 '14 at 20:06
  • Oh sorry, the comparasion test does not works since the limit is zero. But, I cant see what to do with the inequality you gaveme for values of a bigger than 1 – Giiovanna Aug 26 '14 at 20:16
  • I can see what you mean:$ \lim \frac{ln^a(n)}{n} =0 $ for all $a>0$. But, how to show this to generic values of a. I know, by l'Hospital, that it will be $\lim a \frac{log^{a-1}(N)}{N}$, but it seems something recursive. – Giiovanna Aug 26 '14 at 20:23
  • I've found what I was looking for:http://math.stackexchange.com/questions/712613/prove-that-all-values-satisfy-this-expression/712618 – Giiovanna Aug 26 '14 at 20:44
  • It's not endlessly recursive... after $k > a$ iterations you have ${\log^{a - k} n \over n}$, and since the exponent $a - n < 0$ it will only decrease faster than ${1 \over n}$. Alternatively, showing ${\log^a n \over n}$ goes to zero is equivalent to showing its ${1 \over a}$th power ${\log n \over n^{1 \over a}}$ goes to zero, which can be done with a single application of l'Hopital. – Zarrax Aug 27 '14 at 19:22
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If you use the integral test, a substitution of $u=\ln(x)$ changes the integral to the equivalent: $$\int_{\ln(2)}^\infty \frac{e^{u}}{u^a} du.$$ l'Hopital's rule tells us that the integrand blows up to infinity for any $a$ and so the integral diverges.

Now recall that the integral test tells us for a decreasing continuous function $f(x)$, the series $\sum_{n=a}^\infty f(n)$ diverges iff the integral $\int_a^\infty f(x) dx$ diverges.

Thus we can conclude that the series diverges.

Joel
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Since $$ \lim_nn^{-s}(\ln n)^a=0 \quad \forall s\in (0,1], $$ there is some $k\in \mathbb{N}$ such that $$ (\ln n)^a \le n^s \quad \forall n \ge k. $$ There is no loss of generality in assuming that $k \ge 3$. Then we have: $$ \sum_{n=2}^\infty\frac{1}{(\ln n)^a}=\sum_{n=2}^{k-1}\frac{1}{(\ln n)^a}+\sum_{n=k}^\infty\frac{1}{(\ln n)^a} \ge\sum_{n=2}^{k-1}\frac{1}{(\ln n)^a}+\sum_{n=k}^\infty n^{-s} \ge\sum_{n=k}^\infty n^{-s}. $$ Since the series $\sum_{n=k}^\infty n^{-s}$ diverges for $s\in (0,1]$, then the series $\sum_{n=2}^\infty\frac{1}{(\ln n)^a}$ also diverges.

HorizonsMaths
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