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