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Lemma 2 (General difference of power). for any \(x,y \geq 0\) and \(n \in \mathbb {N}\) \begin {align*} x-y = (x^{\frac {1}{n}}-y^{\frac {1}{n}})\sum _{j=0}^{n-1}x^{\frac {j}{n}}y^{\frac {n - j -1}{n}} \end {align*}
Proof. For \(x = y =0\), the result is trivial. For \(y=0\): \begin {align*} x - 0 =(x^{\frac {1}{n}} -0)(x^{\frac {n-1}{n}}+0+...+0) = x \end {align*}
The same reasoning holds for the case where \(y=0\). For the case where \(x,y > 0\). We need observe that: \begin {align*} x-y =y\left (\frac {x}{y}-1\right ) = y\left [\left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{n}-1\right ] \end {align*}
This expression is similar to the result of a geometric series:
\begin {align*} S_n =1+ \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{1} + \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{2}+ ...+ \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{n-1} \\ S_n\left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right ) = \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{1} + \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{2}+ ...+ \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{n}\\ \implies S_n\left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right ) - S_n = \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{n} -1\\ \implies S_n= \dfrac {\left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{n} -1}{\left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right ) -1} \end {align*}
Hence,
\begin {align*} x-y =y\left (\frac {x}{y}-1\right ) = y\left [\left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{n}-1\right ] = y\times S_n \left [ \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right ) -1 \right ] \\ \implies x-y = y\left [ \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right ) -1 \right ]\left [ 1+ \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{1} + \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{2}+ ...+ \left (\frac {x^{\frac {1}{n}}}{y^{\frac {1}{n}}}\right )^{n-1} \right ] \\ \implies x-y = y^{\frac {n-1}{n}}\left [x^{\frac {1}{n}} - y^{\frac {1}{n}} \right ]\left [ y^{\frac {n-1}{n}} + x^{\frac {1}{n}y^{\frac {n-1 -1}{n}}}+... + x^{\frac {n-1}{n}}\right ] \times \frac {1}{y^{\frac {n-1}{n}}}\\ \implies x-y = \left [x^{\frac {1}{n}} - y^{\frac {1}{n}} \right ]\left [ y^{\frac {n-1}{n}} + x^{\frac {1}{n}y^{\frac {n-1 -1}{n}}}+... + x^{\frac {n-1}{n}}\right ] \end {align*}
Now, this is equivalent to: \begin {align*} x-y = (x^{\frac {1}{n}}-y^{\frac {1}{n}})\sum _{j=0}^{n-1}x^{\frac {j}{n}}y^{\frac {n - j -1}{n}} \end {align*}
Hence, for any \(x,y \geq 0\) the above holds true □
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Lemma 3. For any \(a \in (-1,1)\) and \(k,n \in \mathbb {N}\) we have: \begin {align*} \lim \limits _{n \to \infty } n^ka^n =0 \end {align*}
Proof. when \(a=0\), the result is trivial. consider \(0<a<1\). Write: \begin {align*} a = \dfrac {1}{\frac {1}{a}} = \frac {1}{1+b} \end {align*}
Where \(\dfrac {1}{a} > 1\) and \(b = \dfrac {1}{a} - 1 > 0\). Using the binomial theorem:
\begin {align*} a^n &= \dfrac {1}{(1+b)^n} = \dfrac {1}{1 + nb + C^n_2 b^2+...+C^n_{n-1}b^{n-1} +b^n} \\ n^ka^n &= \dfrac {n^k}{1 + nb + C^n_2 b^2+...+C^n_{n-1}b^{n-1} +b^n} \end {align*}
WLOG assume \(n>k\) hence we have: \begin {align*} n^ka^n &= \dfrac {n^k}{1 + nb + C^n_2 b^2+...+C^n_{k+1}b^{k+1}+...+C^n_{n-1}b^{n-1} +b^n} \end {align*}
Observe that: \begin {align*} C^n_{k+1} &= \dfrac {n!}{(n-k -1)!(k+1)!} = \dfrac {n(n-1)(n-2)...(n-k)}{(k+1)!}\\ \implies \dfrac {n^k}{C^n_{k+1}} &= \dfrac {(k+1)!}{n(1-\frac {1}{n})(1-\frac {2}{n})...(1-\frac {k}{n})} \end {align*}
Now because: \begin {align*} C^n_{k+1}b^{k+1} \leq 1 + nb + C^n_2 b^2+...+C^n_{k+1}b^{k+1}+...+C^n_{n-1}b^{n-1} +b^n \end {align*}
We have: \begin {align*} n^ka^n = \dfrac {n^k}{1 + nb + C^n_2 b^2+...+C^n_{n-1}b^{n-1} +b^n} \leq \dfrac {n^k}{C^n_{k+1}} = \dfrac {(k+1)!}{n(1-\frac {1}{n})(1-\frac {2}{n})...(1-\frac {k}{n})}\\ \lim \limits _{n \to \infty }\dfrac {(k+1)!}{n(1-\frac {1}{n})(1-\frac {2}{n})...(1-\frac {k}{n})} = \dfrac {\lim \limits _{n\to \infty }\frac {(k+1)!}{n}}{\lim \limits _{n\to \infty }(1-\frac {1}{n})\lim \limits _{n\to \infty }(1-\frac {2}{n})...\lim \limits _{n\to \infty }(1-\frac {k}{n})} \end {align*}
Because \(\lim \limits _{n\to \infty }\dfrac {k}{n} =0\) for any constant \(k \in \mathbb {N}\): \begin {align*} \dfrac {\lim \limits _{n\to \infty }\frac {(k+1)!}{n}}{\lim \limits _{n\to \infty }(1-\frac {1}{n})\lim \limits _{n\to \infty }(1-\frac {2}{n})...\lim \limits _{n\to \infty }(1-\frac {k}{n})} = \dfrac {(k+1)! \times 0}{(1-0)(1-0)...(1-0)} =0 \end {align*}
Hence, we can indeed bound \(n^ka^n\) above and below. \(n^ka^n\) has a trivial lower bound of \(0\), as here we consider \(0<a<1\): \begin {gather*} 0\leq n^ka^n \leq \dfrac {n^k}{C^n_{k+1}} \\ \lim \limits _{n \to \infty }0 =0 ,\lim \limits _{n \to \infty }\dfrac {n^k}{C^n_{k+1}} =0\\ \implies \lim \limits _{n \to \infty } n^ka^n =0 \text { by squeeze theorem for } a \in (0,1) \end {gather*} For the negative case where \(a \in (-1,0)\) consider: \begin {align*} -|n^ka^n| \leq n^ka^n \leq |n^ka^n| = n^k|a^n| =n^k|a|^n \end {align*}
Because we have already proved that for \(a \in (0,1)\), \(\lim \limits _{n \to \infty }n^ka^n =0\): \begin {align*} \lim \limits _{n \to \infty } n^k|a|^n &=0 \\ \implies \lim \limits _{n \to \infty }-|n^ka^n| &=0 \\ \text {by the squeeze theorem, } \lim \limits _{n \to \infty } n^ka^n &=0 \qquad \text {for } \quad a \in (-1,0) \end {align*}
Hence, for any \(a \in (-1,1)\) and \(k \in \mathbb {N}\) we have: \begin {align*} \lim \limits _{n \to \infty } n^ka^n =0 \end {align*} □