# For which real $s$ does $\sum _{n=1}^{\infty} \frac{1}{n^s}$ converge?

For which values of real $$s$$ does the following converge?

$$\sum _{n=1}^{\infty} \frac{1}{n^s}$$

I have read extensively that the answer is $$s>1$$ but I can't find a simple proof suitable for students not trained to university level mathematics

• The standard proof is by comparison to the corresponding integral. You can also give a pretty elementary proof using Cauchy condensation, which requires no knowledge of calculus: en.wikipedia.org/wiki/Cauchy_condensation_test Dec 10 '20 at 22:54
• @TariqRashid Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) Mar 22 '21 at 18:13

Fix $$s>1$$. For $$k=2,3,\dots$$, one has $$2^k-1=3,7,15,\dots$$ and

\begin{align*} \sum_{n=1}^{2^k-1}\frac{1}{n^s} & = 1 + %% \Big(\frac{1}{2^s}+ \frac{1}{3^s}\Big) + %% \Big(\frac{1}{4^s}+ \frac{1}{5^s} + \frac{1}{6^s}+ \frac{1}{7^s}\Big) + %% %% \cdots + \Big(\frac{1}{(2^{k-1})^s}+ \cdots + \frac{1}{(2^{k}-1)^s}\Big) \\ %% & < 1 + %% \Big(\frac{1}{2^s}+ \frac{1}{2^s}\Big) + %% \Big(\frac{1}{4^s}+ \frac{1}{4^s} + \frac{1}{4^s}+ \frac{1}{4^s}\Big) + %% %% \cdots + \Big(\frac{1}{(2^{k-1})^s}+ \cdots + \frac{1}{(2^{k-1})^s}\Big) \\ %% & = 1 + \frac{2}{2^s} +\frac{4}{4^s}+ %% \cdots + \frac{2^{k-1}}{{2^{k-1}}^s} %% = 1 + 2^{1-s}+ \big(2^{1-s}\big)^2 + %% \cdots + \big(2^{1-s}\big)^{k-1} \\ %% & = \frac{1 - \big(2^{1-s}\big)^{k}}{ 1-2^{1-s}} %% < \frac{1}{ 1-2^{1-s}}<\infty. \\ %% %% \end{align*}

Fix $$s\leq 1$$. For $$k=2,3,\dots$$, one has $$2^k=4,8,16,\dots$$ and

\begin{align*} \sum_{n=1}^{2^k}\frac{1}{n^s} & = 1 + \frac12+ \Big(\frac{1}{3^s}+ \frac{1}{4^s}\Big) + %% \Big(\frac{1}{5^s}+ \frac{1}{6^s} + \frac{1}{7^s}+ \frac{1}{8^s}\Big) + %% %% \cdots + \Big(\frac{1}{(2^{k-1}+1)^s}+ \cdots + \frac{1}{(2^{k})^s}\Big) \\ %% & > 1 + \frac12+ %% \Big(\frac{1}{4^s}+ \frac{1}{4^s}\Big) + %% \Big(\frac{1}{8^s}+ \frac{1}{8^s} + \frac{1}{8^s}+ \frac{1}{8^s}\Big) + %% %% \cdots + \Big(\frac{1}{(2^{k})^s}+ \cdots + \frac{1}{(2^{k})^s}\Big) \\ %% & \geq 1 + \frac12+ %% \Big(\frac{1}{4}+ \frac{1}{4}\Big) + %% \Big(\frac{1}{8}+ \frac{1}{8} + \frac{1}{8}+ \frac{1}{8}\Big) + %% %% \cdots + \Big(\frac{1}{2^{k}}+ \cdots + \frac{1}{2^{k}}\Big) \\ %% & = 1 + \frac12 +\frac12 + \cdots + \frac12 %% = 1 + \frac{k}{2} \rightarrow \infty %% \end{align*}

Here is one way forward that relies only on the Generalized Binomial Theorem and straightforward arithmetic.

For $$s>1$$, clearly we have

\begin{align} n^{1-s}&=\sum_{k=1}^n \left(k^{1-s}-(k-1)^{1-s}\right)\\\\ &=\sum_{k=1}^n k^{1-s} \left(1-\left(1-\frac1k\right)^{1-s}\right)\\\\ \end{align}

Now, the Generalized Binomial Theorem reveals $$\left(1-\frac1k\right)^{1-s}=1-\frac{1-s}{k}+O\left(\frac1{k^2}\right)$$ so that

$$n^{1-s}=\sum_{k=1}^n \frac{1-s}{k^s}+O\left(\frac1{k^{s+1}}\right)\tag1$$

Rearranging $$(1)$$ yields

$$\sum_{k=1}^n \frac{1}{k^s}=\frac{1}{(1-s)n^{s-1}}+\sum_{k=1}^n O\left(\frac1{k^{s+1}}\right)\tag2$$

As $$n\to \infty$$, the first term on the right-hand side of $$(2)$$ goes to zero while the second term is a convergent series since $$s+1>2$$. Therefore, the series on the left-hand side of $$(1)$$ must converge. And we are done.

• Is this not a circular argument? You say the second term in the right-hand-side of (2) converges because $s+1>2$, but that's precisely the type of thing we're trying to prove here. Dec 11 '20 at 0:04
• @Stefan: for exponents $\ge 2$ an elementary argument is possible by comparison to the telescoping series $\sum_{k=1}^{\infty} \frac{1}{k(k+1)}$. The "hard" case is for exponents $<2$. Dec 11 '20 at 0:10
• Might be worth adding this to your answer. Students might not know that. Dec 11 '20 at 0:15
• @stefan Yes. Some students might not have seen the telescoping series "trick" to estimate the for the series $\sum_{n=1}^\infty \frac1{n^2}$. And now Qiaochu's comment serves as the supplemental component to the posted solution. ;-) Dec 11 '20 at 0:37
• Would the downvoter care to comment? Dec 13 '20 at 18:35