<s>[INST] Is it possible to represent the natural number "1" as the sum of p-series in this way? My argument: $$1=(\frac{1}{2})^2+(\frac{1}{3})^2+\cdots+(\frac{1}{2})^3+(\frac{1}{3})^3+\cdots=\sum_{k=2}^\infty (\frac{1}{k})^2+\sum_{k=2}^\infty (\frac{1}{k})^3+\cdots .$$
Explanation) First, for any natural number $n\geq2$, the following holds: 
$$\sum_{k=1}^\infty (\frac{1}{n})^k=(\frac{1}{n})+(\frac{1}{n})^2+(\frac{1}{n})^3+\cdots= \frac{\frac{1}{n}}{1-\frac{1}{n}}=\frac{1}{n-1}.$$ (the infinite geometric series.)
So, we obtain
$$1=(\frac{1}{2})+(\frac{1}{2})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{2})^k,$$
$$\frac{1}{2}=(\frac{1}{3})+(\frac{1}{3})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{3})^k,$$
$$\frac{1}{3}=(\frac{1}{4})+(\frac{1}{4})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{4})^k,$$
and so on.
From above equalities,
$$1=(\frac{1}{2})+(\frac{1}{2})^2+\cdots$$
$$=((\frac{1}{3})+(\frac{1}{3})^2+\cdots)+(\frac{1}{2})^2+(\frac{1}{2})^3+\cdots$$
$$=((\frac{1}{4})+(\frac{1}{4})^2+\cdots)+(\frac{1}{3})^2+(\frac{1}{3})^3+\cdots+(\frac{1}{2})^2+(\frac{1}{2})^3+\cdots$$
$$=\cdots .$$
If $p\ge2$, then p-series absolutely converges. Hence we can change the order of terms in series as follows:
$$1=(\frac{1}{2})^2+(\frac{1}{3})^2+(\frac{1}{4})^2+\cdots$$
$$+(\frac{1}{2})^3+(\frac{1}{3})^3+(\frac{1}{4})^3+\cdots$$
$$+(\frac{1}{2})^4+(\frac{1}{3})^4+(\frac{1}{4})^4+\cdots$$
$$=\sum_{k=2}^\infty (\frac{1}{k})^2+\sum_{k=2}^\infty (\frac{1}{k})^3+\sum_{k=2}^\infty (\frac{1}{k})^4+\cdots.$$
Thus, "1" becomes the sum of p-series(exactly from $k=2$ to infinity).
Is this explanation correct?
 [/INST] So what you are saying is
$1
=\sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m}
$.
Let's check.
$\begin{array}\\
\sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m}
&=\sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \dfrac1{k^m}
\qquad\text{(reverse order of summation)}\\
&=\sum_{k=2}^{\infty}\dfrac{1/k^2}{1-1/k}
\qquad\text{(just a geometric series)}\\
&=\sum_{k=2}^{\infty}\dfrac{1}{k^2-k}
\qquad\text{(multiply num and dec by }k^2)\\
&=\sum_{k=2}^{\infty}\dfrac{1}{k(k-1)}
\qquad\text{(rewrite)}\\
&=\sum_{k=2}^{\infty}(\dfrac1{k-1}-\dfrac1{k})
\qquad\text{(now we can telescope)}\\
&=1\\
\end{array}
$
Yup.
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