# Verify that If $\sum a_n$ converges, then for each $\epsilon>0$ .... $\left|\sum_{n=k}^{\infty} a_n \right| < \epsilon$...

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

Corollary:

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 $\sum \frac{1}{n^p} < \infty$.

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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:

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

• 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 $p > 1$. Aug 7, 2017 at 5:02
• 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. Aug 7, 2017 at 5:03
• 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 $\lim_{a \to \infty} f(a) = 0$. Aug 7, 2017 at 5:05
• @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: $$\lim\limits_{k \rightarrow \infty } \int \frac{1}{n^p}= \lim\limits_{k \rightarrow \infty } \frac{1}{|1-p|k^{|1-p|}} = 0 = \lim\limits_{k \rightarrow \infty} \sum_{n=k}^{\infty} a_n$$
– rei
Aug 7, 2017 at 6:46
• 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. Aug 7, 2017 at 6:50

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.
• thx for the input. Could you elaborate on "$\frac{1}{N}<(\frac{\epsilon}{M})^{1/p}$"?
• 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 $$\frac{1}{N}<\left(\frac{\varepsilon}{M}\right)^{1/p}.$$ Aug 7, 2017 at 6:02
• @JohnGriffin What you wrote is not correct, check again. The goal is to say it converges iff $p > 1$ Aug 7, 2017 at 6:38