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In the question Self-Contained Proof that $\sum\limits_{n=1}^{2k+1}\frac{1}{n^p}$ Converges for $p>1$, joriki made a proof to show that when $p>1$, $\sum\limits_{n=1}^{2k+1}\frac{1}{n^p}$ converges:

We can bound the partial sums by multiples of themselves:

$$\begin{eqnarray} S_{2k+1} &=& \sum_{n=1}^{2k+1}\frac{1}{n^p}\\ &=& 1+\sum_{i=1}^k\left(\frac{1}{(2i)^p}+\frac{1}{(2i+1)^p}\right)\\ &<&1+\sum_{i=1}^k\frac{2}{(2i)^p}\\ &=&1+2^{1-p}S_k\\ &<&1+2^{1-p}S_{2k+1}\;. \end{eqnarray}$$

Then solving for $S_{2k+1}$ yields

$$S_{2k+1}<\frac{1}{1-2^{1-p}}\;,$$

and since the sequence of partial sums is monotonically increasing and >bounded from above, it converges.

I understand how this proof show that when $p>1$, $\sum\limits_{n=1}^{2k+1}\frac{1}{n^p}$ converges and I know that when $p\leq1$, this series diverges. However, I cannot understand why when $p<1$, the series become less than a negative value shown by the proof. How does the proof get to this step as the original series can never be less than $0$? And does this proof also mean that when $p\leq1$, this series diverges because being less than a negative automatically shows divergence?

lxywizard
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  • The sentence "is less than a series that is not existed" is confusing me. Can you explain what doesn't exist? – Michael Burr Jan 04 '18 at 03:18
  • @MichaelBurr I just deleted this sentence since mostly I am confused about the "less than a negative value" part. – lxywizard Jan 04 '18 at 03:26
  • You are getting an error when $p<1$ because you are dividing an inequality by a negative number. Note that $1-2^{1-p} <0$ when $p<1$. The last step in the argument is false but the previous step is correct. – Kavi Rama Murthy Jan 04 '18 at 06:26
  • @KaviRamaMurthy Thank you! It is exactly the reason that made me feel confused. – lxywizard Jan 04 '18 at 08:44

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We have $S_{2k+1}<1+2^{1-p}S_{2k+1},$ implying $$S_{2k+1}(1-2^{1-p})<1.$$ However, if $p\leq 1$ then $1-2^{1-p}\leq 0,$ and then the inequality above yields no useful information, and does $not$ imply the last inequality of your Q. An inequality of the form $AB<1$ with $A=S_{2k+1}>0$ and $B\leq 0$ tells us nothing more about $A.$

$S_{2k+1}$ is always positive.

A different method is needed to show divergence when $p\leq 1.$

The Cauchy Condensation Test gives a quick answer to convergence or divergence for all p.

DanielWainfleet
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