To prove the convergence of the p-series

$$\sum_{n=1}^{\infty} \frac1{n^p}$$

for $p > 1$, one typically appeals to either the Integral Test or the Cauchy Condensation Test.

I am wondering if there is a self-contained proof that this series converges which does not rely on either test.

I suspect that any proof would have to use the ideas behind one of these two tests.


10 Answers 10


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


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

Eight years after I posted this, Zach Teitler pointed out that this proof was already published in $1979$ by Teresa Cohen and William J. Knight in Convergence and Divergence of $\sum_{n=1}^{\infty} 1/n^p$, Mathematics Magazine, $52(3)$, $1979$, p. $178$ (accessible version, DOI link).

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    $\begingroup$ @Glen: I never saw this; I just made it up when I saw the question :-) $\endgroup$
    – joriki
    Mar 28, 2011 at 8:23
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    $\begingroup$ This is a really slick proof. I get a glimpse of the Condensation Test in it, but some students might like this argument better. $\endgroup$ Mar 28, 2011 at 10:10
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    $\begingroup$ @joriki: This is great. When I teach Calc II, I teach series before integration. (It gets the hardest stuff out of the way first.) But, that means I can't use the integral test, which is most often used for p-series. With your proof in hand, I don't need the integral test at all! $\endgroup$ Mar 28, 2011 at 14:09
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    $\begingroup$ Schlömilch's condensation test says that if $a_n$ is a nonnegative and decreasing sequence, then $$ \sum_{n=1}^{\infty} a_n \text{ converges} \qquad \text{if and only if} \qquad \sum_{k=1}^{\infty} (n_{k+1} - n_{k}) a_{n_k} \text{ converges} $$ where $n_k$ is any strictly increasing sequence of positive integers such that $$ \frac{n_{k+1} - n_{k}}{n_{k} - n_{k-1}} $$ is bounded. Note that Cauchy's condensation test takes $n_k = 2^k$. $\endgroup$
    – admchrch
    Mar 28, 2011 at 16:34
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    $\begingroup$ Very nice proof. You can you use the same kind of trick to extend the $\zeta$ function on the critical strip by checking that $(1-2^s) \, \zeta(s) = \eta(s)$ where the Dirichlet $\eta$ function is defined by $\eta(s) = \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^s}$. $\endgroup$
    – Joel Cohen
    May 15, 2011 at 0:10

My personal favourite is a variant of a common proof that the harmonic series diverges: we get $$\sum_{n=2^k}^{2^{k+1}-1}\frac1{n^p}\le2^k\cdot\frac1{2^{kp}}=2^{(1-p)k}.$$ because the sum has $2^k$ terms of which the first is the largest. Now sum this over $k$ to get $$\sum_{n=1}^{\infty}\frac1{n^p}\le\sum_{k=0}^\infty2^{(1-p)k}=\frac1{1-2^{1-p}}<\infty$$ since $2^{1-p}<1$.

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    $\begingroup$ Yeah, I know I am late to this particular party, but I figured this one also deserved to be in the list of solutions. $\endgroup$ Apr 11, 2012 at 17:03
  • $\begingroup$ hi, I know this question is from 2012, but I am wondering how do you get to the first line, where $\sum 1/n^p$ .... , thanks ! $\endgroup$
    – Smarties
    Oct 15, 2018 at 21:43
  • $\begingroup$ @MathAvengers As I explain just below that line, by multiplying the number of terms by the largest one. (The inequality is actually strict, except when $k=0$.) $\endgroup$ Oct 17, 2018 at 13:18
  • $\begingroup$ I just checked: this was already explained on the page for roughly a year when you posted it. $\endgroup$
    – Did
    Jan 19, 2019 at 16:18
  • $\begingroup$ @Did When I posted my answer, I clearly had not noticed the same proof buried in the fourth paragraph of your answer, or I would at least have made a reference to it. Sorry about that, but it really isn't obvious unless you read carefully. $\endgroup$ Jan 20, 2019 at 14:42

FWIW, the properties of the generalized harmonic series $\sum \frac{1}{n^p}$ can be used to prove another convergence criterion (which goes under the name of second Cauchy's convergence criterion, as far as I remember).

The criterion is the following:

Let $(a_n)$ be a sequence of positive numbers. If:

$$\lim_{n\to \infty} \frac{\ln \frac{1}{a_n}}{\ln n} =L>1 \tag{1}$$

then the series $\sum a_n$ converges. On the other hand, if:

$$\lim_{n\to \infty} \frac{\ln \frac{1}{a_n}}{\ln n} =l<1 \tag{2}$$

then the series $\sum a_n$ diverges.

The proof is very simple. The criterion remains valid even if one replaces $\displaystyle \lim_{n\to \infty}$ with $\displaystyle \liminf_{n\to \infty}$ in (1) and with $\displaystyle \limsup_{n\to \infty}$ in (2).

  • $\begingroup$ This is very nice. It is an analog of the root test, whose proof relies on comparison with a geometric series. The proof here relies on comparison with a $p$-series (i.e., a generalized harmonic series). $\endgroup$
    – admchrch
    Mar 28, 2011 at 16:02

This is the most direct and elementary way I know how to prove the result, although it only works for powers in the range $[0,1] \cup [2,\infty)$ which is exactly the uninteresting set, and Joriki's answer is much better regardless. I had already written this, and perhaps somebody finds it useful.

First we consider $p=1$. Set $x_n$ equal to the greatest power of 2 less than $\frac{1}{n}$. That is, $$(x_n) = (\frac{1}{2}, \frac{1}{4},\frac{1}{4}, \frac{1}{8}, \frac{1}{8},\frac{1}{8},\frac{1}{8},\dots).$$

Note that $$\sum_{i=1}^{\infty} x_n = \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1}x_j\Big) = \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1} \frac{1}{2^{i+1}}\Big) = \sum_{i=1}^{\infty} \frac{1}{2}, $$ which diverges, and that $x_n < \frac{1}{n}$, which proves $\sum_{n=1}^{\infty} \frac{1}{n}$ also diverges.

If $p = 2$, let $x_n = \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ if $n>1$ and 1 if $n=1$.

Observe that $$\sum_{n=1}^{\infty} x_n = 1 + \sum_{n=2}^{\infty}\Big(\frac{1}{n-1} - \frac{1}{n}\Big) = 1 + 1 - \lim_{n\rightarrow\infty}(1/n) = 2,$$ and so in particular this series converges.

Now $x_n > \frac{1}{n \cdot n} = \frac{1}{n^2},$ so the series $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ converges as well.

Finally if $p>2$ then since $\frac{1}{n^p} < \frac{1}{n^2}$, the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges, and if $0<p<1$ we have $\frac{1}{n^p} > \frac{1}{n}$, and so $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges.

  • $\begingroup$ This is the way I knew for the longest time, before taking a course in Real Analysis and learning the proper Techniques and Results $\endgroup$ Mar 14, 2023 at 10:29

For every $p>1$, one can reduce the problem to the convergence of some geometric series. Then, either one takes the latter for granted or one proves it by an elementary argument. More details below.

Let $N(k)$ denote the integer part of $a^k$, where the real number $a>1$ is defined by $a^{p-1}=2$. The sum of $1/n^p$ over $n$ between $N(k)$ and $N(k+1)$ is at most the number of terms $N(k+1)-N(k)$ times the greatest term $1/N(k)^p$. This contribution behaves like $$ (a^{k+1}-a^k)/a^{kp}\sim(a-1)a^{k(1-p)}=(a-1)2^{-k}. $$ The geometric series of general term $2^{-k}$ converges hence the original series converges.

There is a variant of this, where one considers the slabs of integers limited by the powers of $2$. The contribution of slab $k$ is then at most the number of terms $2^k$ times the greatest term $1/2^{kp}$. This reduces the problem to the convergence of the geometric series of general term $b^k$, where $b=2^{1-p}<1$.

Finally, as everybody knows, a simple proof of the convergence of the geometric series of general term $b^k$, for every $b$ in $(0,1)$ say, is to note that the partial sums $s_k$ are such that $s_0=1$ and $s_{k+1}=1+bs_k$, and to show by induction on $k$ that $s_k\le1/(1-b)$ for every $k\ge0$.

Edit The so-called variant above shows, using $b=2^{1-p}$, that the sum of the series is at most $1/(1-b)=2^p/(2^p-2)$. But the contribution of slab $k$ is also at least the number of terms $2^k$ times the smallest term $1/2^{(k+1)p}$, hence the sum of the series is at least $1/(2^p-2)$. Finally the sum of the series is between $1/(2^p-2)$ and $2^p/(2^p-2)$.

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    $\begingroup$ are you familiar with Cauchy's Condensation Test? Because this really looks pretty similar to it. $\endgroup$ Mar 28, 2011 at 10:11
  • $\begingroup$ @Pete: I am. $ $ $\endgroup$
    – Did
    Mar 28, 2011 at 10:18
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    $\begingroup$ okay, had I seen your quite erudite comments on the subject to Joriki's answer earlier I would not have asked. :) $\endgroup$ Mar 28, 2011 at 13:38
  • $\begingroup$ @Pete: No problem. In fact your question made me re-think about this whole condensation test thing, which led to the comments on @joriki's answer. So, in the end, I benefitted from your question. :-) $\endgroup$
    – Did
    Mar 28, 2011 at 14:01

Here is another proof:

By Bernoulli

$$\frac{(n+1)^p}{n^p} \geq 1+ \frac{p}{n}=\frac{n+1}{n}+\frac{p-1}{n}$$

Multiplying by $n$ and dividing by $(n+1)^{p}$ you get

$$\frac{1}{n^{p-1}}\geq \frac{1}{(n+1)^{p-1}} + \frac{p-1}{(n+1)^p}$$


$$\frac{p-1}{(n+1)^p} \leq \frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}} $$


$$(p-1) \sum_{n=2}^{m+1} \frac{1}{n^p}=(p-1) \sum_{n=1}^m \frac{1}{(n+1)^p} \leq \sum_{n=1}^m \frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}}\,.$$

Since the last sum is telescopic you get

$$(p-1) \sum_{n=2}^{m+1} \frac{1}{n^p} \leq 1- \frac{1}{(m+1)^{p-1}}$$

  • $\begingroup$ Bernoulli's inequality is wrong for $1\lt p\lt2$. For $p\geqslant2$, simpler methods exist. $\endgroup$
    – Did
    Jan 30, 2012 at 6:21
  • $\begingroup$ @DidierPiau Thanks, forgot about that :) It's fixed now, I think, anyhow, Bernoulli for real powers uses analysis.... $\endgroup$
    – N. S.
    Jan 30, 2012 at 6:38
  • $\begingroup$ Yes, basically this is a comparison with the integral of $x\mapsto1/x^p$ on suitable intervals, which explains why the sum is telescopic. $\endgroup$
    – Did
    Jan 30, 2012 at 6:46
  • $\begingroup$ Actually, this is similar to my answer to another question. Bernoulli on integer exponents can be proven using induction, and further extended to rational exponents, also using induction. $\endgroup$
    – robjohn
    Sep 2, 2023 at 19:08

I suppose what joriki proved could be phrased as $$|\zeta(s)| \leq \frac{1}{1-2^{1-\sigma}} \ \ \ \ \ \forall \sigma >1 , \ \ s=\sigma+it$$ Instead of taking the sum in his second line over $\mathbb{Z}_2,$ one could do this over $\mathbb{Z}_q$ for any integer $q>1$ to get the nice inequality $$\sum_{n > q}\frac{1}{n^\sigma} \leq q^{1-\sigma} \ \ \zeta(\sigma) \ \ \ \ \ \ \ \ \forall \sigma>1, \ q \geq 2, q \in \mathbb{Z}$$ Now given any real $x>2$ we can use $q=1+[x]$ in the case that $x$ is not an integer to get $$\zeta_{x}(\sigma):=\sum_{n > x}\frac{1}{n^\sigma} \leq x^{1-\sigma} \ \ \zeta(\sigma) +O(\frac{1}{x^{\sigma}})\ \ \ \ \ \ \ \ \forall \sigma>1, \ x \geq 2, x \in \mathbb{R}$$ and we can therefore phrase this as $$\zeta(s)=\zeta_{x}(s)+O(\frac{x^{1-\sigma}}{1-\sigma}+\frac{1}{x^{\sigma}}) \ \ \ \ \forall \sigma>1, \ x \geq 2, x \in \mathbb{R}$$ uniformly in the stated regions. This inequality might be of some interest regarding the truncated zeta function.


We first show that for $k\geq 1$ $$ \sum_{n=1}^\infty\frac{1}{n(n+1)\dotsb(n+k)}=\frac{1}{k\times k!}\tag{0}. $$ This result combined with the limit comparison test will yield the fact that $\sum n^{-p}<\infty$ for $p>1$.

Indeed, the series in (0) telescopes as can be seen from the partial fraction decomposition $$ \frac{1}{n(n+1)\dotsb(n+k)}=\frac{1}{k}\left( \frac{1}{n(n+1)\dotsb(n+k-1)}-\frac{1}{(n+1)\dotsb(n+k)} \right) $$ whence $$ \sum_{n=1}^m\frac{1}{n(n+1)\dotsb(n+k)}=\frac{1}{k}\left( \frac{1}{k!}-\frac{1}{(m+1)\dotsb(m+k)} \right). $$ Letting $m\to\infty$ yields the result.

Next observe that for $p>1$ we have that $$ \frac{1}{n^p}\sim\frac{1}{n(n+1)\dotsb(n+p-1)}. $$ Apply the limit comparison test and we are done.


Let $S(n)$ = $ \Sigma_{1}^{n} \frac{1}{n^p}$


$S(2n)$ = $S(n)$ + $\frac{1}{{(n+1)}^p} $ + $\frac{1}{{(n+2)}^p} $ + $\frac{1}{{(n+3)}^p} $ +$\frac{1}{{(n+4)}^p} $ ...........+ $\frac{1}{{(n+n)}^p} $

Let $\Delta S$ = $S(2n) - S(n)$

then $\frac{n}{{(2n)}^p}$ $\leq$ $\Delta S$ $\leq$ $\frac{n}{(n+1)^p} $

By sandwitch theorem we see that if $p > 1$, $\lim_{n \rightarrow \inf}$ $\Delta S = 0$

so that as n tends to infinity $S(n) = S(2n) = S(4n) ....$

So that series converges


$S(2n) - S(n) \leq \frac{n}{(n+1)^p} \leq n^{1-p} $

$S(2^{k+1}n) - S(2^{k}n) \leq \frac{2^{k+1}n - 2^{k}n}{(2^kn+1)^p} \leq \frac{2^{k+1}n - 2^{k}n}{(2^kn)^p} \leq n^{1-p}2^{k(1-p)}$

Summing above equation for k = 0 to m, we get

$S(2^{m+1}n) - S(n) \leq n^{1-p}(\frac{1-2^{(m+1)(1-p)}}{1-2^{(1-p)}})$

Hence $\lim_{n \to \infty} R(n) =\lim_{m \to \infty}S(2^{m+1}n) - S(n) =0 $

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    $\begingroup$ I like the way this post begins, but alas this answer is wrong. In particular, the statement, "as $n$ tends to infinity, $S(n) = S(2n) = S(4n) = \cdots$" does not make sense. $\endgroup$
    – Srivatsan
    Jan 30, 2012 at 6:02
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    $\begingroup$ @Srivatsan, First of all thanks for feedback, as I was eager to know if it is correct. I understand your point so thanks for pointing out that. I will try to improve this proof. $\endgroup$
    – chatur
    Jan 30, 2012 at 6:34
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    $\begingroup$ BTW I did not downvote your post (although I was contemplating it). As I said, I do see a good idea in the post, but the details are all awry. $\endgroup$
    – Srivatsan
    Jan 30, 2012 at 6:40
  • $\begingroup$ @chatur, answer is really similar to answer by N.S $\endgroup$
    – alekhine
    Aug 2, 2013 at 9:36
  • $\begingroup$ @Srivatsan, can you please look at the question math.stackexchange.com/questions/457418/… $\endgroup$
    – alekhine
    Aug 2, 2013 at 9:39

We have. $S=\sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^p$ $p>1$ S = $1+\left(\frac{1}{2}\right)^p+\left(\frac{1}{3}\right)^p+\left(\frac{1}{4}\right)^p+\left(\frac{1}{5}\right)^p+\left(\frac{1}{6}\right)^p+\left(\frac{1}{7}\right)^p+\left(\frac{1}{8}\right)^p+\cdots \cdot$. clearly, $$ \begin{aligned} S & <1+\left(\frac{1}{2}\right)^P+\left(\frac{1}{2}\right)^p+\left(\frac{1}{4}\right)^p+\left(\frac{1}{4}\right)^p+\left(\frac{1}{4}\right)^p+\left(\frac{1}{4}\right)^p+\cdots \\ & =1+\frac{2}{(2)^p}+\frac{4}{(4)^p}+\frac{8}{(8)^p}+... \end{aligned} $$ $$ =1+\left(\frac{1}{2}\right)^{p-1}+\left(\frac{1}{2}\right)^{2(p-1)}+\left(\frac{1}{2}\right)^{3(p-1)}+\cdots \cdots $$ which is sum of an infinite geometric progression. $$ \Rightarrow s<\frac{1}{1-\left(\frac{1}{2}\right)^{p-1}} \text { if }\left|\frac{1}{2}\right|_{\text {which is true since p>1}}^{p-1}<1 $$ $\Rightarrow S$ is bounded above. Also $S$ is an increasing sequence $\Rightarrow S$ converges to $\operatorname{sup}(S)$.

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    $\begingroup$ My first answer.pls ignore latex errors. $\endgroup$ Aug 18, 2023 at 4:35

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