# Thread: Discuss the convergence of the series

1. ## 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 $\sum_{n=1}^{\infty} \frac{3n}{4(n+1)}$

Solution :

Ratio test :

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

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

Comparison test :

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

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

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

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

What should i do !!!!!

2. Originally Posted by flintstone
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 $\sum_{n=1}^{\infty} \frac{3n}{4(n+1)}$

[/COLOR][/I][/B]!!!!!
You have got to be kidding me? Try comparing the series to 1

3. Originally Posted by Drexel28
You have got to be kidding me? Try comparing the series to 1
Sorry i didnt get u ......which test should i use !!!!!!!

4. Originally Posted by flintstone
Sorry i didnt get u ......which test should i use !!!!!!!
Try the limit comparison test.

5. Originally Posted by Drexel28
Try the limit comparison test.
sir i have tried the limit comparison test .........please see my first post for the steps .....

6. Originally Posted by flintstone
sir i have tried the limit comparison test .........please see my first post for the steps .....
$\lim_{n\to\infty}\frac{\frac{3n}{4(n+1)}}{1}=\frac {3}{4}$

Look up limit test for divergence.

7. Originally Posted by Drexel28
$\lim_{n\to\infty}\frac{\frac{3n}{4(n+1)}}{1}=\frac {3}{4}$

Look up limit test for divergence.
u have taken $u_n = \frac{3n}{4(n+1)}$ and $v_n = 1$ is that correct ??........are we allowed to take $v_n =1$(just 1)

8. Originally Posted by flintstone
u have taken $u_n = \frac{3n}{4(n+1)}$ and $v_n = 1$ is that correct ??........are we allowed to take $v_n =1$(just 1)
Surely.

9. What Drexel is saying is that in any series,

If $\lim_{n \to \infty}{a_n} \neq 0$,

then the series diverges.

Look at the $n^{th}$ term in your series.

$a_n = \frac{3n}{4(n + 1)}$

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

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

$= \frac{3}{4}\lim_{n \to \infty}\left(1 - \frac{1}{n + 1}\right)$

$= \frac{3}{4}\cdot 1$

$= \frac{3}{4}$.

Since this is not 0, the series diverges.