Showing a series converges

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irobaf

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Jul 20, 2022
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Dear all,
can someone help me and say me if he agrees with what i did?
Here is my problem:

Let $\mathbf{Y} \in \mathbb{R}^n$, $p,p_0,m \in \mathbb{N}$ such that $p-p_0 \geq 2$, $\Gamma(\cdot)$ the gamma function and $H$ a projection matrix onto a linear subspace $M \subset \mathbb{R}^n$.
I have to show that the following series converges (it is not necessary true): ∑k=0∞(∣∣HY∣∣2∣∣Y∣∣)2ke−m2−2k(2k)2k(m+2k)m2+2kk!∣∣Y∣∣mΓ(k+p−p02)\sum_{k=0}^\infty \left(\frac{||H\mathbf{Y}||}{2||\mathbf{Y}||}\right)^{2k} \frac{e^{-\frac{m}{2}-2k}(2k)^{2k}(m+2k)^{\frac{m}{2}+2k}}{k!||\mathbf{Y}||^{m}\Gamma\left(k + \frac{p-p_0}{2}\right)}.

To do so, i used the root test:
R=lim⁡k→∞[(∣∣HY∣∣2∣∣Y∣∣)2ke−m2−2k(2k)k(m+2k)m2+kk!∣∣Y∣∣mΓ(k+p−p02)]1kR =\lim_{k\rightarrow \infty} \left[\left(\frac{||H\mathbf{Y}||}{2||\mathbf{Y}||}\right)^{2k} \frac{e^{-\frac{m}{2}-2k}(2k)^{k}(m+2k)^{\frac{m}{2}+k}}{k!||\mathbf{Y}||^{m}\Gamma\left(k + \frac{p-p_0}{2}\right)}\right]^{\frac{1}{k}}=∣∣HY∣∣2e−24∣∣Y∣∣2lim⁡k→∞2k(m+2k)e−m2k(m+2k)m2k(k!)1kΓ(k+p−p02)1k∣∣Y∣∣m2k= \frac{||H\mathbf{Y}||^2e^{-2}}{4||\mathbf{Y}||^2} \lim_{k\rightarrow \infty} 2k(m+2k)\frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{(k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}}||\mathbf{Y}||^{\frac{m}{2k}}}=∣∣HY∣∣2e−24∣∣Y∣∣2lim⁡k→∞e−m2k(m+2k)m2k∣∣Y∣∣m2k2k(m+2k)(k!)1kΓ(k+p−p02)1k= \frac{ ||H\mathbf{Y}||^2e^{-2} }{ 4||\mathbf{Y}||^2 } \lim_{k\rightarrow \infty} \frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{||\mathbf{Y}||^{\frac{m}{2k}}} \frac{ 2k(m+2k) }{ (k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} }
Then, I used lim⁡k→∞e−m2k(m+2k)m2k∣∣Y∣∣m2k=1\lim_{k\rightarrow \infty} \frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{||\mathbf{Y}||^{\frac{m}{2k}}} = 1
With the Stirling formula $\Gamma(z + b) \sim \sqrt{2\pi}e^{-z}z^{z+b-1/2}$ for $b\in \mathbb{R}$, i have:


lim⁡k→∞(k!)1/k=elim⁡k→∞1kln⁡(Γ(k+1))\lim_{k \rightarrow \infty} (k!)^{1/k} = e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln(\Gamma(k+1))} =elim⁡k→∞1kln⁡(2πe−k−1(k+1)k+1/2)= e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\sqrt{2\pi}e^{-k-1}(k+1)^{k+1/2}\right)} =elim⁡k→∞ln⁡(2π)k−k+1k+k+1/2kln⁡(k+1)= e^{\lim_{k\rightarrow\infty}\frac{\ln\left(\sqrt{2\pi}\right)}{k} - \frac{k+1}{k} + \frac{k+1/2}{k}\ln(k+1)} =lim⁡k→∞(k+1)= \lim_{k\rightarrow\infty} (k+1) and
lim⁡k→∞Γ(k+p−p02)1k=elim⁡k→∞1kln⁡(Γ(k+p−p02))\lim_{k\rightarrow\infty}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} = e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\Gamma\left(k+\frac{p-p_0}{2}\right)\right)} =elim⁡k→∞1kln⁡(2πe−k(k)k+p−p0−12)= e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\sqrt{2\pi}e^{-k}(k)^{k+\frac{p-p_0-1}{2}}\right)} =elim⁡k→∞ln⁡(2π)k−kk+kkln⁡(k)+p−p0−12kln⁡(k)= e^{\lim_{k\rightarrow\infty}\frac{\ln\left(\sqrt{2\pi}\right)}{k} - \frac{k}{k} + \frac{k}{k}\ln(k)+ \frac{p-p_0-1}{2k}\ln(k)} =elim⁡k→∞ln⁡(k)(ln⁡(2π)kln⁡(k)−1ln⁡(k)+1+p−p0−12k)= e^{\lim_{k\rightarrow\infty}\ln(k)\left(\frac{\ln\left(\sqrt{2\pi}\right)}{k\ln(k)} - \frac{1}{\ln(k)} + 1 + \frac{p-p_0-1}{2k}\right)} =lim⁡k→∞k= \lim_{k\rightarrow\infty} k.


Then we have:

lim⁡k→∞2k(m+2k)(k!)1kΓ(k+p−p02)1k\lim_{k\rightarrow\infty} \frac{ 2k(m+2k) }{ (k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} }=lim⁡k→∞2k(m+2k)(k+1)k=4= \lim_{k\rightarrow\infty} \frac{ 2k(m+2k) }{ (k+1)k} = 4.
Finally, we then have:
R=∣∣HY∣∣2e−24∣∣Y∣∣2⋅1⋅4=∣∣HY∣∣2e−2∣∣Y∣∣2R = \frac{ ||H\mathbf{Y}||^2e^{-2} }{ 4||\mathbf{Y}||^2 } \cdot 1 \cdot 4 = \frac{ ||H\mathbf{Y}||^2e^{-2} }{ ||\mathbf{Y}||^2 }
because ∣∣HY∣∣∣∣Y∣∣=∣∣Y∣∣−∣∣(In−H)Y∣∣∣∣Y∣∣≤1\frac{||H\mathbf{Y}||}{||\mathbf{Y}||} = \frac{||\mathbf{Y}|| - ||(I_n-H)\mathbf{Y}||}{||\mathbf{Y}||} \leq 1 we have $R \leq e^{-2} < 1$.
 
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