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Let $p \in \mathbb{R}$ and $p>0$. When the series

$ \sum_{n=1}^{\infty} \frac{\ln(1+n^p)}{n^p}$

is convergence (dependent from parametr $p$) ?

Of course, $\lim_{n \to \infty} \frac{\ln(1+n^p)}{n^p} = 0$. It is true, because $p>0$.

Next, I tried different convergence tests (especially D'Alembert's criterion, comparison test, Dirichlet's test) but without success.

I will grateful for you hints and help.

Thomas
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4 Answers4

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First note that a positive series either converges or diverges (to $+\infty$).

Now, let $0 < p \leq 1$. Then, $$\sum_{n=1}^{\infty} \frac{\log(1+n^p)}{n^p} \geq \sum_{n=1}^{\infty} \frac{\log(1+n^p)}{n} \geq \sum_{n=k}^{\infty} \frac{1}{n},$$where $k$ is the smallest positive integer such that $\log(1+k^p) \geq 1$. Since the lower bound diverges, the original series diverges.

Now, let $p = 1+\epsilon$ for some $\epsilon>0$. We have $$\sum_{n=1}^{\infty} \frac{\log(1+n^p)}{n^p} \leq \sum_{n=1}^{k-1} \frac{\log(1+n^p)}{n^p} + \sum_{n=k}^{\infty} \frac{n^{\epsilon/2}}{n^{1+\epsilon}} = \sum_{n=1}^{k-1} \frac{\log(1+n^p)}{n^p} + \sum_{n=k}^{\infty} \frac{1}{n^{1+\epsilon/2}},$$ where this $k$ is the smallest positive integer such that $\log(1+k^p) \leq k^{\epsilon/2}$. The first term in the upper bound is a constant, while the second term converges. Hence, the upper bound converges, and thus our series is convergent.

Therefore, the series converges for every $p>1$ and diverges for any $0<p\leq 1$.

Lord Soth
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    Ok, but how we can prove that exist this $k$ , so that $\log(1+k^p) \le k^{\epsilon/2}$ ? – Thomas Jul 25 '13 at 18:36
  • You may take $f(n) = \frac{\log(1+n^p)}{n^{\epsilon/2}}$ and show that $\lim_{n\rightarrow\infty}f(n) = 0$ (I guess you have already done something similar). This implies $f(n) =\frac{\log(1+n^p)}{n^{\epsilon/2}} \leq 1$ for every sufficiently large $n$ (or for $n\geq k$ where $k$ is the guy we are looking for). – Lord Soth Jul 25 '13 at 18:50
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You can use the integral test. We consider the integral

$$\int_{1}^{\infty}\frac{\ln(1+x^p)}{x^p}dx= \lim _{x\rightarrow \infty }{\frac {\ln \left( 1+{x}^{p} \right) }{ \left( 1-p \right) {x}^{p-1}}}+{\frac {\ln \left( 2 \right) }{-1+p}}+\frac{p}{p-1} \int _{1}^{\infty }\!{\frac {1}{ \left( 1+{x}^{p } \right) }}{dx}.$$

Now, you can see that, the above limit diverges if $0<p<1$ together with the last integral, since

$$ \int _{1}^{\infty }\!{\frac {1}{ \left( 1+{x}^{p } \right) }}{dx} \sim \int _{1}^{\infty }\!{\frac {1}{ {x}^{p } }}{dx} $$

which makes the whole integral diverges and hence the series diverges. On the other hand, the limit is $0$ and the integral converges for $p>1$ which implies that the series converges.

Community
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Mhenni Benghorbal
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We have

$$\frac{\ln\left(1+n^p\right)}{n^p}\sim_\infty\frac{\ln\left(n^p\right)}{n^p}=p\frac{\ln(n)}{n^p}$$

So

  • if $0\leq p\leq 1$ then $\frac{\ln(n)}{n^p}\geq \frac{1}{n^p}$ and the given series is divergent.
  • if $p>1$ then pick $p'$ such that $1<p'<p$ and we have $\frac{\ln(n)}{n^p}=_\infty o\left(\frac{1}{n^{p'}}\right)$ and the given series is convergent.
0

Try

$$p\ln(n)=\ln(n^p)<\ln(1+n^p) < \ln(2n^p) =\ln(2) +p\ln(n)<2p\ln(n) $$

DonAntonio
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david
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