Prove that $$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$ converges if $p>1$ and diverges when $0<p\leqslant1$.

My attempt:

Since the terms of the series $\sum_{n=0}^{\infty} \frac1n$ are monotonically decreasing and positive, I applied Cauchy's Condensation theorem, i.e, if $\sum a(n)$ and $\sum2^na(2^n)$ will converge and diverge together. Applying this I got, $\sum\left(1-\frac{1}{2^n}\right)^p$, but I am stuck now. Don't know how to proceed further.

  • $\begingroup$ Could you please improve the readibility of your question by using the $\LaTeX$ support? It should be $\sum_{n\geq 1}\frac{1}{n H_n^p}$, I guess. $\endgroup$ Commented Dec 9, 2015 at 11:16
  • $\begingroup$ I tryed to decode your text. Please check if it is correct $\endgroup$
    – AndreasT
    Commented Dec 9, 2015 at 11:18
  • $\begingroup$ @AndreasT, yes question is correct and now it is much more readable. Thanks. $\endgroup$
    – Utkarsh
    Commented Dec 9, 2015 at 11:23
  • 1
    $\begingroup$ If $p>0$ then the sum is more than $\sum\frac1n$ which diverges. $\endgroup$
    – Empy2
    Commented Dec 9, 2015 at 11:25
  • $\begingroup$ @Michael, The series given is ∑1/n(1+1/2+....+1/n)^p and not ∑1/n, so I assume there must be some different concept to solve it. $\endgroup$
    – Utkarsh
    Commented Dec 9, 2015 at 11:34

3 Answers 3


On one hand since $\frac {1}{n}\le \frac 1n\left(1+\frac 12+\ldots+\frac 1n\right)$ it is not difficult to see that for $p>0$ we have

$$\sum_{n=0}^{\infty} \frac {1}{n^p}\le\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$ so the divergence statement from Riemann series which diverges for $p\le 1$

On the other hand it follows from Hardy inequality that: for $p>1$ $$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p\le \left(\frac{p}{p-1}\right)^p\sum_{n=0}^{\infty} \frac {1}{n^p}<\infty$$ the converges blatantly follows from the Riemann series on the right hand side which converges only for $p>1.$


By Riemann sums, we have $\log n\lt\sum\limits_{i=1}^n\frac1i\lt1+\log n$
This leads to $\sum \frac1n(\log n)^p$, which, by another Riemann sum, can be compared with $\int_1^n\frac1x(\log x)^pdx=\int_0^{\log x}y^pdp$

  • $\begingroup$ How does it shows that ∫(y^p)dp converges for p>1 and diverges for p≤1? $\endgroup$
    – Utkarsh
    Commented Dec 10, 2015 at 5:33

In $$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$

$$1+\frac 12+\ldots+\frac 1n=O(\log(n))$$

so , given sum equals

$$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p=\sum_{n=0}^{\infty}\left(\frac{\log(n)}{n}\right)^p$$

Here, $O(n)\gg O(\log(n))$ ,

so convergence of given sum depends on convergence of $$\sum_{n=0}^{\infty}\left(\frac{1}{n}\right)^p$$

which converges if $p>1$ and diverges when $0<p\leq1$.

For case $p=1$

given sum equals $$\sum_{n=0}^{\infty}\left(\frac{\log(n)}{n}\right)$$ which is divergent as


converges if $p>1$


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