# Thread: convergence of sequence via inequation

1. ## convergence of sequence via inequation

Hi! I would like to proof that $\displaystyle a_n:=\ln(n!)+n-(n+\frac{1}{2}) \ln(n)$ converges. I already have that
$\displaystyle (n+\frac{1}{2}) \ln(n+\frac{1}{2})-n \leq \ln(n!) \leq (n+1) \ln(n+1)-n$ or equivalently
$\displaystyle (n+\frac{1}{2}) \ln(1+\frac{1}{2n}) \leq \ln(n!)+n-(n+\frac{1}{2}) \ln(n) \leq (n+\frac{1}{2}) \ln(1+\frac{1}{n})+ \frac{1}{2} \ln(n+1)$

I have also calculated the difference $\displaystyle a_n - a_{n+1}=(n+\frac{1}{2}) \ln(1+\frac{1}{n})-1$ but i do not know how to use my inequations.

2. Originally Posted by gammafunction
Hi! I would like to proof that $\displaystyle a_n:=\ln(n!)+n-(n+\frac{1}{2}) \ln(n)$ converges. I already have that
$\displaystyle (n+\frac{1}{2}) \ln(n+\frac{1}{2})-n \leq \ln(n!) \leq (n+1) \ln(n+1)-n$ or equivalently
$\displaystyle (n+\frac{1}{2}) \ln(1+\frac{1}{2n}) \leq \ln(n!)+n-(n+\frac{1}{2}) \ln(n) \leq (n+\frac{1}{2}) \ln(1+\frac{1}{n})+ \frac{1}{2} \ln(n+1)$

I have also calculated the difference $\displaystyle a_n - a_{n+1}=(n+\frac{1}{2}) \ln(1+\frac{1}{n})-1$ but i do not know how to use my inequations.

I may be wrong, but I feel like your inequality is useless: the lower bound tends to 1/2, while the upper bound diverges to $\displaystyle +\infty$ (because of the last log).
However, if you know series, here's a way: find an asymptotic equivalence for $\displaystyle a_{n+1}-a_n$ in order to show that $\displaystyle \sum_n (a_{n+1}-a_n)$ converges. Try to see why this is enough.
Exactly that was my problem! In order to proof that $\displaystyle a_n$ converges i was given the hint to use $\displaystyle \ln(n!)=\ln(1)+...+\ln(n)$ and the inequations $\displaystyle \int \limits_{k-\frac{1}{2}}^{k+\frac{1}{2}} \ln(x) dx \leq \ln(k) \leq \int \limits_k^{k+1} \ln(x) dx$. Do you know how i could use this somehow?