# Thread: Test the convergence of the series

1. ## Test the convergence of the series

Question : Test the convergence of the series $\sum_{n = 1}^\infty \frac{3n}{2(n + 1)}$

2. use the divergence test to show that the limit as n->infinity =3/2 so it diverges

3. ## Can u please provide me with the starting few steps

Originally Posted by Sam1111
use the divergence test to show that the limit as n->infinity =3/2 so it diverges

Can u please provide me with the starting few steps

4. The divergence test states that if the limit of 3n/(2(n+1)) doesnt = 0 then the series diverges.

So take the limit of 3n/(2(n+1)) by dividing everything by the highest power in the denominator. Then evaluate this letting n->infinity.

5. ## Could u please tell me if its correct or no

Originally Posted by Sam1111
The divergence test states that if the limit of 3n/(2(n+1)) doesnt = 0 then the series diverges.

So take the limit of 3n/(2(n+1)) by dividing everything by the highest power in the denominator. Then evaluate this letting n->infinity.

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

$\sum_{n=1}^{ \infty} \frac{3n}{2(n+1)}$

$\lim_{n \to infty} \frac{3n}{2(n+1)}$ = $\lim_{n \to infty} \frac{3n}{n (2+ \frac{1}{n})}$ = $\lim_{n \to \infty} \frac{3}{2 + \frac{1}{n}}$ = $\frac{3}{2}$

$since \sum_{n = 1}^{ \infty} u_n \ne 0$ then series is divergent

$\therefore$ $\lim_{n \to \infty} \frac{3n}{2(n+1)}$ is divergent

6. Originally Posted by zorro
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

$\sum_{n=1}^{ \infty} \frac{3n}{2(n+1)}$

$\lim_{n \to infty} \frac{3n}{2(n+1)}$ = $\lim_{n \to infty} \frac{3n}{n (2+ \frac{1}{n})}$ = $\lim_{n \to \infty} \frac{3}{2 + \frac{1}{n}}$ = $\frac{3}{2}$

$since \sum_{n = 1}^{ \infty} u_n \ne 0$ then series is divergent

$\therefore$ $\lim_{n \to \infty} \frac{3n}{2(n+1)}$ is divergent
No no no ...

A series $\sum_{i=1}^{\infty} a_i$ converges only if $\lim_{i \to \infty}a_i=0$. In your case $a_i \to 3/2 \ne 0$.

CB

7. ## I didnt get u

Originally Posted by CaptainBlack
No no no ...

A series $\sum_{i=1}^{\infty} a_i$ converges only if $\lim_{i \to \infty}a_i=0$. In your case $a_i \to 3/2 \ne 0$.

CB

Thats what i am saying the series is divergent ...........since $\sum_{n = 1}^{ \infty} u_n \ne 0$

U got me there mite i am confused now

8. Originally Posted by zorro
Thats what i am saying the series is divergent ...........since $\sum_{n = 1}^{ \infty} u_n \ne 0$

U got me there mite i am confused now
No since it's divergent because $\lim_{n \to \infty}u_n \ne 0$ which is not the same thing as you posted.

To see this consider:

$\sum_{n = 1}^{ \infty} 0.1^n=\frac{1}{9}$

So if $u_n=0.1^n$ we have $\sum_{n = 1}^{ \infty} u_n \ne 0$ and yet the series converges, but we do have $\lim_{n \to \infty}u_n = 0$ since this is a necessary condition for the series to converge.

CB

9. Originally Posted by CaptainBlack
No since it's divergent because $\lim_{n \to \infty}u_n \ne 0$ which is not the same thing as you posted.

To see this consider:

$\sum_{n = 1}^{ \infty} 0.1^n=\frac{1}{9}$

So if $u_n=0.1^n$ we have $\sum_{n = 1}^{ \infty} u_n \ne 0$ and yet the series converges, but we do have $\lim_{n \to \infty}u_n = 0$ since this is a necessary condition for the series to converge.

CB

So u mean to say i should have posted as $\lim_{n \to \infty}u_n \ne 0$ that is why the series is divergent is tht correct

10. Originally Posted by zorro
So u mean to say i should have posted as $\lim_{n \to \infty}u_n \ne 0$ that is why the series is divergent is tht correct
Yes.

CB