for $1<p<2$, prove the p-series is convergent without concerned with integral and differential knowledge and geometry series

for $1<p<2$, prove the p-series: $\sum_{n=1}^{\infty}n^{-p}$ is convergent.

please use Cauchy Rule (edit: that is, by showing directly that the sequence of partial sums is a Cauchy sequence) instead of concerned with integral and differential knowledge.

I have seen a proof before, without concern any tests about series. But I forgot it.

I had learned the condensation test. But the proof I had seen is not constructed by proved the condensation test, more precisely, not by shown every specify partial sums of a $p$-series is bounded in the corresponding items of a geometry series.

Edit. So much grateful to user6312 and Didier for your warmhearted doings.

For some reason however, I had to bother you two although I would wish have not to. User6312’s second answer seem as a proof without importing differential or integral point but, the base idea, showing EACH $p$-series (here $p=2^{-k}$) was a telescope series SURELY could be done, came from the “middle point theorem”. What do you think? For Didier’s proof, like user6312’s denote, a geometry series behind it, since you had to recurring to get the general term’s formula of $A_n$.

I recalled the ever seen proof a little last night. That is like assemble from your two answers:

$\forall p\in(1,2)$, $\exists k\in\mathbb{N}$ so that $2^{-k}<p-1$, then show the cutted-partial-sum (that is, summary from $N$ to $N+m$) of $2^{-k}$-series yields the SUM (multiply a constant factor) of $2^{1-k}$-series (if it’s convergent), $\ldots\,$. At last (after finite steps), show the $2^{-1}$-series is a telescope series. Since the cutted-partial-sum of given $p^{-k}$-series yields the cutted-partial-sum of $2^{-k}$-series is obvious (to consider orresponding term), we have done. But the detail I haven’t thought through over clearly. Might someone write it down?

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Here is a version of the Cauchy Condensation Test: If $(a_n)$ is a decreasing sequence of positive terms, then
$$\sum_{n=1}^\infty a_n$$ converges precisely if
$$\sum_{k=0}^\infty 2^k a_{2^k}$$
converges. This may be a little hard to read: $a_{2^k}$ is the $2^k$-th term of the sequence.

Let $a_n=1/n^p$. Then
$$a_{2^k}=\frac{1}{2^{kp}}$$
So the Cauchy Condensation Test says that your series converges iff
$$\sum_{k=0}^\infty \frac{2^k}{2^{kp}}$$
converges.
The above series can be rewritten as
$$\sum_{k=0}^\infty \frac{1}{(2^{p-1})^k}$$
Since $p>1$, $2^{p-1}>1$. Let $r=1/(2^{p-1})$. Note that $0<r<1$.
Then the series obtained from Cauchy Condensation is
$$\sum_{k=0}^\infty r^k$$
You will recognize this as a convergent geometric series, and we are finished.
With a more complex expression than $1/n^p$, the series obtained from the Cauchy Condensation Test might need further processing.

Edit A large chunk is added below to the original answer, since recently the OP made it clear that what meant by the “Cauchy test” was not Cauchy Condensation, but proving that the partial sums form a Cauchy sequence.

We want to show that for any $\epsilon>0$, we can find an $N=N(\epsilon)$, such that for any $m$ and $n$, if $N<m<n$, then
$$(m+1)^{-p}+(m+2)^{-p} +\cdots + (n)^{-p} <\epsilon$$
For any $m$, let $2^k$ be the largest integer such that $2^k < m$. It is enough to show that if $k$ is large enough, the longer sum
$$(2^k)^{-p}+(2^k+1)^{-p} + (2^k+2)^{-p} +(2^k+3)^{-p} +\cdots$$
is less than $\epsilon$. (We are summing from $2^k$ to “infinity” to get rid of the pesky and useless $n$. We can if we wish stop the summing at the smallest power of $2$ which is $>n$.)

So look at the displayed sum above. Break this up into the sum
$(2^{k})^{-p}+ (2^{k}+1)^{-p}+\cdots + (2^{k+1}-1)^{-p}$, plus the sum $(2^{k+1})^{-p}+ (2^{k+1}+1)^{-p}+\cdots + (2^{k+2}-1)^{-p}$, plus the sum $(2^{k+2})^{-p}+ (2^{k+2}+1)^{-p}+\cdots +(2^{k+3}-1)^{-p}$, and so on.

Look first at
$$(2^{k})^{-p}+(2^{k}+1)^{-p}+\cdots + (2^{k+1}-1)^{-p}$$
Each term in the sum is $\le (2^k)^{-p}$, and the sum has $2^k$ terms, so
it adds up to something $\le (2^k)^{1-p}$.

Similarly, the next chunk adds up to something $\le (2^{k+1})^{1-p}$, and so on. Thus our entire sum is less than or equal to (actually less than)
$$(2^k)^{1-p}+ (2^{k+1})^{1-p} +
(2^{k+2})^{1-p}+\cdots$$
This sum can be rewritten as
$$(2^{1-p})^{k}+ (2^{1-p})^{k+1} +
(2^{1-p})^{k+2}+\cdots$$

Note that $2^{p-1}>1$. So the sum above is a geometric series with first term $(2^{1-p})^k$ and common ratio $2^{1-p}$. We can find an explicit expression for the sum, which is
$$\frac{(2^{1-p})^k}{2^{p-1}-1}$$
From this expression it is easy to obtain a value of $2^k$ that will make the sum $<\epsilon$.

The following is an alternate answer, based on telescoping series, not geometric series. The procedure makes it easy to show that the partial sums are Cauchy, by making it easy to find an explicit bound.

We present first something the OP did not ask for: a proof that the series
$$\frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2}+ \cdots $$
converges.

Note that
$$\frac{1}{(n+1)^2} <\frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}$$
It follows that
$$\sum_{k=m+1}^{n} \frac{1}{k^2} <\frac{1}{m}$$
by the usual “telescoping series” argument. We can then use the “Cauchy sequence” criterion to conclude that our original series converges. And then, if $p > 2$, we can use the obvious inequality $1/n^{p} \le 1/n^2$ to deal with $p$-series with $p>2$. In order to do this, we need to know what $n^p$ means. But there is no difficulty if $p$ is an integer, or even just rational.

So far, simple and familiar. We now ask: can the same telescoping series argument be used to deal with the case $1<p<2$ that was asked about? It turns out that the answer is yes.

We will need to work with inequalities involving $1/n^{p}$. This is fine if we have already introduced the general exponential function, but it becomes problematical otherwise. So we will start with the $p$-series with $p=1+1/2$. Then we will look at the $p$-series with $p=1+1/4$.

We will find, for $p=3/2$, an inequality analogous to the one we used in the case $p=2$. So look at $1/(n+1)^{3/2}$. It would be nice if this were less than $1/n^{1/2} -1/(n+1)^{1/2}$, for then nothing much would change.
That is almost what happens.
$$\frac{1}{n^{1/2}} -\frac{1}{(n+1)^{1/2}}=\frac{(n+1)^{1/2}-n^{1/2}} {n^{1/2}(n+1)^{1/2}}$$

Multiply “top” and “bottom” by $(n+1)^{1/2}+n^{1/2}$. We obtain
$$\frac{1}{n^{1/2}(n+1)^{1/2}((n+1)^{1/2} +n^{1/2})}$$
and this is clearly bigger than
$$\frac{1}{2(n+1)^{3/2}}$$

We conclude that
$$\frac{1}{(n+1)^{3/2}}<\frac{2}{n^{1/2}} -\frac{2}{(n+1)^{1/2}}$$
and now the telescoping series argument works.

We deal next with $p=5/4=1+1/4$.
The natural candidate to compare with $1/(n+1)^{5/4}$ is $1/n^{1/4}-1/(n+1)^{1/4}$. Bring this expression to a common denominator. We obtain
$$\frac{(n+1)^{1/4}-n^{1/4}}{n^{1/4}(n+1)^{1/4}}$$

Rationalize the numerator by multiplying top and bottom by
$(n+1)^{1/4}+n^{1/4}$, and then by $(n+1)^{1/2}+n^{1/2}$. We obtain
$$\frac{1}{n^{1/4}(n+1)^{1/4}((n+1)^{1/4}+n^{1/4})((n+1)^{1/2}+n^{1/2})}$$
This is clearly bigger than
$$\frac{1}{4(n+1)^{5/4}}$$
We conclude that
$$\frac{1}{(n+1)^{5/4}}<\frac{4}{n^{1/4}} -\frac{4}{(n+1)^{1/4}}$$
and again the telescoping argument works.

For $p=9/8=1+1/8$, the same idea works, except that we will need to multiply by $3$ terms to rationalize the numerator. We can continue in this way forever.

So the conclusion is that if $p=1+1/2^k$, the $p$-series converges. For more general $p>1$, just find an integer $k$ such that $1+1/2^k <p$, and use some basic properties of the (general) exponential function.

Alternately, we could use standard inequalities involving the exponential function to show that (for $p>1$), $1/(n+1)^p <C(1/n^{p-1} -1/(n+1)^{p-1})$ for some constant $C=C(p)$, then use the telescoping series argument. Perhaps it is this sort of argument that the OP recalls seeing. This is a guess, but the problem specifies $1<p<2$, as if $p \ge 2$ were already dealt with earlier. We chose to deal with $3/2$, $5/4$, and so on to keep the calculations at a more concrete level.

One can show that the sequence of partial sums is Cauchy, as follows. Introduce the notations
$$
a_n=1/n^p,\qquad A_N=\sum_{n=0}^Na_n.
$$
As in the condensation test, we begin with the powers of $2$, hence we consider
$$
B_k=\sum_{n=2^k}^{2^{k+1}-1}a_n.
$$
Bounding every $a_n$ in $B_k$ by $a_{2^k}=1/2^{kp}$ and the number of terms by $2^k$, one gets
$$
B_k\le2^k/2^{kp}=r^k,\qquad r=2^{-(p-1)}.
$$
Since $p>1$, $r<1$ and since one knows the sum of a series of geometric general term, one has for any $K$,
$$
\sum_{k=K}^{+\infty}B_k\le\sum_{k=K}^{+\infty}r^k=cr^K,\qquad c=1/(1-r).
$$
Now, fix $N$ and compare the partial sums $A_N$ and $A_{N'}$, for any $N'\ge N$. Of course $A_N\le A_{N'}$. Furthermore, picking an integer $K$ such that $2^K\le N+1$, one gets
$$
A_{N'}-A_{N}=\sum_{n=N+1}^{N'}a_n\le\sum_{n=N+1}^{+\infty}a_n\le\sum_{n=2^K}^{+\infty}a_n=\sum_{k=K}^{+\infty}B_k\le cr^K.
$$
We are ready to show that the sequence $(A_N)$ is Cauchy. Let $\varepsilon>0$. Pick $K$ such that $cr^K\le\varepsilon$ (this is possible since $c$ is a constant and $r<1$) and let $N_\varepsilon=2^K$. Then for every $N\ge N_\varepsilon$ and every $N'\ge N$,
$$
0\le A_{N'}-A_{N}\le\varepsilon,
$$
that is, the sequence $(A_N)_N$ is Cauchy.

Edit Here is a variant which might satisfy the OP. First, since every $a_n$ is positive, the series converges if and only if the sequence $(A_N)$ is bounded. Second, $A_{2N}$ is made of even numbered terms and odd numbered terms. The sum of the even numbered terms is $A_N/2^p$ because $a_{2n}=a_n/2^p$ for every $n$. Likewise, $a_{2n+1}\le a_{2n}$ for every $n\ge1$ hence the sum of the odd numbered terms is
$$
1+\sum_{n=1}^{N-1}a_{2n+1}\le1+\sum_{n=1}^{N-1}a_{2n}=1+A_{N-1}/2^p\le1+A_N/2^p.
$$
This yields $A_{2N}\le1+2A_N/2^p$, which implies by recursion over $N\ge1$ that $A_N\le1/(1-2^{1-p})$ for every $N\ge1$. The sequence $(A_N)$ is bounded, hence the series is summable.