How to compute that
$\lim_{n\to\infty}\Big(\ln(\Gamma(n+x+1)) - \ln(\Gamma(n+1)) - a\ln(n)\Big)$,
for $x>0$, $a\geq 0$ and
$\lim_{x\to\infty}\Big(\sum_{k=1}^\infty \frac{x+a}{k^2+k(x+a)} - \sum_{k=1}^\infty \frac{x}{k^2+kx} - \frac{a}{x}\Big)$,
for $a\geq 0$ . Anyone can help me? thanks.
On
By using the definition of the Gamma function $ \Gamma(n) = (n-1)! $ we can rewrite the above equation as: $$ \lim_{n\rightarrow\infty}(\ln(\Gamma(n+x+1)-\ln(\Gamma(n+1)-a\ln(n)) = \lim_{n\rightarrow\infty}(\ln(n+x)!)-\ln(n!)-\ln(n^a)) = $$ $$ = \lim_{n\rightarrow\infty} \ln\Big(\frac{(n+x)!}{n!n^a}\Big)$$ The term $ (n+x)! $ can be rewritten as: $$ (n+x)! = (n+x)\cdot (n+x-1)\cdot...\cdot(n+1)\cdot n! $$ the limit now becomes: $$ \lim_{n\rightarrow\infty} \ln\Big(\frac{(n+x)\cdot (n+x-1)\cdot...\cdot(n+1)}{n^a}\Big) $$ In the product of the numerator the highest degree will be of order $x$: $$ (n+x)\cdot (n+x-1)\cdot...\cdot(n+1) = n^x+a_1n^{x-1}+...+a_xn+a_{x+1} $$ Where the $a$'s are just constants. When this product is divided by $n^a$ we will have three cases to consider:
Case $1$: $ a>x $. In this case the fraction: $$ \frac{n^x+a_1n^{x-1}+...+a_xn+a_{x+1}}{n^a} \rightarrow 0_+$$ as $n\rightarrow \infty$ and thus the limit: $$\lim_{n\rightarrow\infty} \ln\Big(\frac{(n+x)\cdot (n+x-1)\cdot...\cdot(n+1)}{n^a}\Big) = -\infty$$ Case $2$ : $ a<x $. In this case the fraction: $$ \frac{n^x+a_1n^{x-1}+...+a_xn+a_{x+1}}{n^a} \rightarrow +\infty$$ as $n\rightarrow \infty$ and thus the limit: $$\lim_{n\rightarrow\infty} \ln\Big(\frac{(n+x)\cdot (n+x-1)\cdot...\cdot(n+1)}{n^a}\Big) = +\infty$$ Case $3$: $ a=x $. In this case the fraction: $$ \frac{n^x+a_1n^{x-1}+...+a_xn+a_{x+1}}{n^a} \rightarrow 1$$ as $n\rightarrow \infty$ and thus the limit: $$\lim_{n\rightarrow\infty} \ln\Big(\frac{(n+x)\cdot (n+x-1)\cdot...\cdot(n+1)}{n^a}\Big) = 0$$
For the second problem the two sums converge by the ratio test and thus you can add them together and you get: $$\sum_{k=1}^\infty \frac{x+a}{k^2+k(x+a)} - \sum_{k=1}^\infty \frac{x}{k^2+kx} = \sum_{k=1}^\infty \frac{a}{(k+x+a)(k+x)} = $$ $$ = \sum_{k=1}^\infty \frac{1}{(k+x+a)}-\frac{1}{(k+x)}$$ This is a telescoping series and all the terms after $k$ surpasses $a$ will cancel out. The part that survives are the value for the first $k < a$ values: $$\sum_{k=1}^\infty \frac{1}{(k+x+a)}-\frac{1}{(k+x)} = \sum_{k=1}^{a-1}-\frac{1}{k+x}$$ As this sum is finite and in the limit $x\rightarrow\infty$ it will go to zero and the limit is also going to 0.
Hint:
Use Stirling's approximation that $$\ln(\Gamma(n+r+1))\sim\frac{1}{2}\ln(2\pi n) +(n+r)\left(\ln(n+r)-1\right)$$ When $n\to\infty$