Verify Corollary below for the convergent series, with $p>1$, $$\sum \frac{1}{n^p}$$


If $\sum a_n$ converges, then for each $\epsilon>0$ there is an integer $K$ such that $$\left|\sum_{n=k}^{\infty} a_n \right| < \epsilon$$ if $k \geq K$ that is $$ \lim\limits_{k \rightarrow \infty } \sum_{n=1}^{\infty} a_n = 0$$

As $\frac{1}{n^p}$ is a decreasing positive function of for an interval $[1, \infty)$, with $n\geq k \geq 0$ and $p>1$, $$\int_{k}^{\infty} f(n) = \lim\limits_{t \rightarrow \infty} \left[ \frac{1}{1-p} n^{1-p} \right]_k^t = \lim\limits_{t \rightarrow \infty} \left[ \frac{1}{1-p} t^{1+p} - \frac{1}{1-p} k^{1-p} \right]= \frac{1}{|1-p|k^{|1-p|} } $$ since $k$ is finite and $p>1$. As $\int f < \infty$, it follows that $\sum \frac{1}{n^p} < \infty$.


By comparison test, since $ \frac{1}{n^p} \geq \frac{1}{(n+1)^p} => \sum \frac{1}{(n+1)^p} < \infty$

Since both converges, then there are for every $\epsilon_n>0$ and $\epsilon_{n+1}>0$, two integers $K_n$ and $K_{n+1}$ such that

$$\left|\sum_{n=k}^{\infty} a_n \right| = \left|\frac{1}{n^p} \right|< \epsilon_n$$ $$\left|\sum_{n=k}^{\infty} a_{n+1} \right|= \left|\frac{1}{(n+1)^p} \right| < \epsilon_{n+1}$$

It follows that $$ \left|\frac{1}{(n+1)^p} \right| < \left|\frac{1}{n^p} \right| < \epsilon_{max \{\epsilon_{n+1},\epsilon_n\}} $$

that is (as when $n \rightarrow \infty, n+1 \rightarrow \infty)$

$$ \lim\limits_{k \rightarrow \infty } \sum_{n=k}^{\infty} a_n = 0 $$

Is my argumentation correct/appropriate? What can be improve? what to do out of this limit at the end of the corrollary? Much appreciated for your input/help.

RE-EDIT: To address the veracity of $ \lim\limits_{k \rightarrow \infty } \sum_{n=k}^{\infty} a_n = 0 $ Is this appropriate to state the following:

$$\lim\limits_{k \rightarrow \infty } \int \frac{1}{n^p}= \lim\limits_{k \rightarrow \infty } \frac{1}{|1-p|k^{|1-p|}} = 0 = \lim\limits_{k \rightarrow \infty} \sum_{n=k}^{\infty} a_n $$

  • $\begingroup$ Yes it is correct. In short words $|\sum_{n=a}^b n^{-p}| \le \int_{a-1}^{b+1} x^{-p} dx$ and $\int_1^\infty x^{-p}dx= \lim_{b \to \infty} \frac{b^{1-p}-1}{1-p}$ converges for $p > 1$. $\endgroup$
    – reuns
    Aug 7, 2017 at 5:02
  • $\begingroup$ I assume you mean $$\lim _{n\to\infty}\sum_{n=1}^\infty a_n = 0 \quad\mbox{or}\quad \lim_{k\to\infty}\sum_{k=1}^\infty a_k= 0$$ which is true for any convergent series. $\endgroup$
    – AlvinL
    Aug 7, 2017 at 5:03
  • $\begingroup$ Except that $\sum_{n=a}^\infty c_n$ doesn't mean anything if it doesn't converge, so the definition is $\sum_{n=1}^\infty c_n$ converges iff it is Cauchy, which means for every $b$, $|\sum_{n=a}^b c_n| \le f(a)$ with $f$ finite and $\lim_{a \to \infty} f(a) = 0$. $\endgroup$
    – reuns
    Aug 7, 2017 at 5:05
  • $\begingroup$ @reuns To address the veracity of $ \lim\limits_{k \rightarrow \infty } \sum_{n=k}^{\infty} a_n = 0 $ Is this appropriate to state the following: $$\lim\limits_{k \rightarrow \infty } \int \frac{1}{n^p}= \lim\limits_{k \rightarrow \infty } \frac{1}{|1-p|k^{|1-p|}} = 0 = \lim\limits_{k \rightarrow \infty} \sum_{n=k}^{\infty} a_n $$ $\endgroup$
    – rei
    Aug 7, 2017 at 6:46
  • $\begingroup$ Unclear. You only need to show $\lim_{N \to \infty} |\sum_{n=1}^N a_n| < \infty$. Here $a_n \ge 0$ and $a_n \le \int_{n-1}^n x^{-p}dx$ thus $|\sum_{n=1}^N a_n| \le 1+\int_1^N x^{-p}dx$ which is easy to handle. $\endgroup$
    – reuns
    Aug 7, 2017 at 6:50

1 Answer 1


The question is asking for you to verify that $$\lim_{k\to\infty}\sum_{n=k}^\infty \frac{1}{n^p}=0,$$ where $p>1$. We know that $M:=\sum_{n=1}^\infty \frac{1}{n^p}$ exists. Given $\varepsilon>0$, choose a natural number $N$ such that $$\frac{1}{N}<\left(\frac{\varepsilon}{M}\right)^{1/p}.$$ Whenever $k>N$ we have $$ \sum_{n=k}^\infty\frac{1}{n^p} = \frac{1}{(k-1)^p}\sum_{n=1}^\infty\frac{1}{n^p} = \frac{M}{(k-1)^p} \leq \frac{M}{N^p}<\varepsilon. $$ This completes the proof.

  • $\begingroup$ thx for the input. Could you elaborate on "$\frac{1}{N}<(\frac{\epsilon}{M})^{1/p}$"? $\endgroup$
    – rei
    Aug 7, 2017 at 5:51
  • 1
    $\begingroup$ Do you mean to ask why we know that such an $N$ exists? It is a consequence of the fact that $$\left(\frac{M}{\varepsilon}\right)^{1/p}$$ is a fixed real number. Thus we can choose a natural number $N$ so large as to satisfy $$N>\left(\frac{M}{\varepsilon}\right)^{1/p},$$ and consequently $$\frac{1}{N}<\left(\frac{\varepsilon}{M}\right)^{1/p}.$$ $\endgroup$ Aug 7, 2017 at 6:02
  • $\begingroup$ @JohnGriffin What you wrote is not correct, check again. The goal is to say it converges iff $p > 1$ $\endgroup$
    – reuns
    Aug 7, 2017 at 6:38
  • $\begingroup$ Since it states "verify the corollary below for the convergent series...", one is to assume that we already know that the series converges, and we just wish to check this consequence. $\endgroup$ Aug 7, 2017 at 14:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .