2

Find all $p\geq 0$ such that the following series converges $\sum_{k = 1}^{\infty} \frac{1}{k\log^p(k+1)}$.

Proof: the general term for the series is $\frac{k^p}{k^p\log^p(k+1)^n} = \frac{1}{k\log^p(k+1)^n}$.

By comparison, $\frac{1}{\log^pn}\leq \frac{1}{\log^p(n+1)} $. And it's convergent when $p>1$ thus $\sum_{k = 1}^{\infty} \frac{1}{k\log^p(k+1)}$ is convergent when $p>1$ and divergent when $0 < p < 1$.

Can anyone please verify this? Any suggestion would help. I was trying to use the integral test, but I don't know how to do it. Thank you

Olivier Oloa
  • 122,789
user848242
  • 61

3 Answers3

3

Recall that, by the integral test, for a monotonically decreasing non-negative function $f$, you have $$ \begin{align} \int_1^\infty f(x)dx\leq\sum_{k=1}^\infty f(k)\leq f(1)+\int_1^\infty f(x)dx. \end{align} $$ Here, if you put $\displaystyle x \mapsto f(x):=\frac{1}{x (\ln(x+1))^p}, \: p>0$, you get that $\sum_{k=1}^\infty f(k) $ and $\int_1^\infty f(x)dx $ either both converge or both diverge.

  1. Let $p$ be such that $p>1$.

    You may write $$ \begin{align} &\int_1^{\infty} \frac{1}{x (\ln(x+1))^p}\, dx\\\\ &=\int_1^{\infty} \frac{x+1}{x}\frac{\frac{1}{(x+1)}}{(\ln(x+1))^p}\, dx\\\\ &=\left.\frac{x+1}{x}\left(-\frac{1}{p-1}\frac{1}{(\ln(x+1))^{p-1}}\right)\right|_1^{\infty} -\frac{1}{p-1}\int_1^{\infty} \frac{1}{x^2}\frac{1}{(\ln(x+1))^{p-1}} dx\\\\ &=\frac{2}{(p-1) (\ln 2)^{p-1}} - \frac{1}{p-1}\int_1^{\infty} \frac{1}{x^2}\frac{1}{(\ln(x+1))^{p-1}} dx \end{align} $$ the latter integral is convergent since, the potential problem being as $x \to +\infty$ and we have $$ \frac{1}{x^2}\frac{1}{(\ln(x+1))^{p-1}} \leq \frac{1}{x^2}, \quad x\geq2, $$ your series is thus convergent in this case.

  2. Let $p$ be such that $0<p<1$.

    Since, for $x \to +\infty$, you have $$ \frac{1}{x (\ln(x+1))^p} \sim \frac{1}{x (\ln x)^p}$$ then, for $M$ sufficiently great, you get $$ \int_M^{+\infty}\!\! \frac{1}{x (\ln(x+1))^p}\:dx \sim \int_M^{+\infty}\!\! \frac{1}{x (\ln x)^p} \:dx$$ the latter integral is divergent since $$ \int_M^{+\infty}\!\! \frac{1}{x (\ln x)^p} dx= \int_M^{+\infty}\!\! \frac{(\ln x)'}{(\ln x)^p}\:dx=\left.\frac{(\ln x)^{1-p}}{1-p}\right|_M^{+\infty}=+\infty$$ and your series is divergent in this case.

Olivier Oloa
  • 122,789
1

By Cauchy condensation test all gets reduced to the study of $$\sum_{k = 1}^{\infty} \frac{1}{k^p}$$ whence we get the conclusions according to the values of $p$.

Q.E.D.

user 1591719
  • 44,987
0

Let $f (k) = \frac{1}{k\log^p(k+1)}$. When $g (k) = 2^k f (2^k)$, we have $$\sum f (k) \leqslant \sum g (k) \leqslant 2 \sum f (k),$$ so if one converges, the other converges too. We have $$g (k) = \frac {1} {(k + 1)^{p \log 2}}.$$ Then $\sum g (k)$, so $\sum f (k)$ converges when $p > 1/\log 2$.