Question : Test the convergence of the series $\displaystyle \sum_{n = 1}^\infty \frac{3n}{2(n + 1)}$
I am not so good with the convergent / divergent series so i dont know whether i have done correctly or no please check and let me know
$\displaystyle \sum_{n=1}^{ \infty} \frac{3n}{2(n+1)}$
$\displaystyle \lim_{n \to infty} \frac{3n}{2(n+1)}$ = $\displaystyle \lim_{n \to infty} \frac{3n}{n (2+ \frac{1}{n})}$ = $\displaystyle \lim_{n \to \infty} \frac{3}{2 + \frac{1}{n}}$ = $\displaystyle \frac{3}{2}$
$\displaystyle since \sum_{n = 1}^{ \infty} u_n \ne 0$ then series is divergent
$\displaystyle \therefore$ $\displaystyle \lim_{n \to \infty} \frac{3n}{2(n+1)}$ is divergent
No since it's divergent because $\displaystyle \lim_{n \to \infty}u_n \ne 0$ which is not the same thing as you posted.
To see this consider:
$\displaystyle \sum_{n = 1}^{ \infty} 0.1^n=\frac{1}{9}$
So if $\displaystyle u_n=0.1^n$ we have $\displaystyle \sum_{n = 1}^{ \infty} u_n \ne 0$ and yet the series converges, but we do have $\displaystyle \lim_{n \to \infty}u_n = 0$ since this is a necessary condition for the series to converge.
CB