Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$

Question

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.

Answer 1

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.

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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).

Answer 2

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 $2^{1-p}<1$.

Answer 3

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).

Answer 4

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.

Answer 5

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 $2^p/(2^p-2)$.

Answer 6

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

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

or

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

Thus,

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

Since the last sum is telescopic you get

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

Answer 7

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.

Answer 8

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.

Answer 9

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

then

$S(2n)$ = $S(n)$ + $\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

EDIT:

$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 $

Answer 10

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 $\operatorname{sup}(S)$.