Discuss the convergence of the series

*I have tried the ratio test and the comparison test but still i am unable to get the result *

Question : Discuss the convergence of the infinite series $\displaystyle \sum_{n=1}^{\infty} \frac{3n}{4(n+1)}$

** Solution** :

** Ratio test** :

$\displaystyle \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{3(n+1)}{4(n+2)}. \frac{4(n+1)}{3n} = \lim_{n \to \infty} \frac{(n+1)^2}{(n+2)n}$ now dividing nu and de by $\displaystyle n^2$ we get

$\displaystyle \lim_{n \to \infty} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{1+\frac{2}{n}} = 1 (Ratio test failed )$

** Comparison test** :

$\displaystyle \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{3n}{4(n+1)} $

Here $\displaystyle u_n = \frac{3n}{4(n+1)} ; v_n = \frac{1}{n}$

$\displaystyle \lim_{n \to \infty } \frac{u_n}{v_n} = \lim_{n \to \infty } \frac{3}{4(n+1)}$ = $\displaystyle \frac{3}{4} \lim_{n \to \infty } \frac{3}{4(n+1)}$ dividing the nu and de by n weget

$\displaystyle \frac{3}{4} \lim_{n \to \infty } \frac{\frac{3}{n}}{4+\frac{1}{n}}$ = $\displaystyle \frac{3}{4} . 0 $ = 0

*What should i do *!!!!!