# Thread: Test the Convergence of the series

1. ## Test the Convergence of the series

This question is from my exam paper which i had appeared last semister i am trying to solve this but dont know the right answer.

Q: Test the convergence of the series

∑ 3n/2(n+1)
n=1

Please replies with explanation if possible .......

2. Originally Posted by varunnayudu
This question is from my exam paper which i had appeared last semister i am trying to solve this but dont know the right answer.

Q: Test the convergence of the series

∑ 3n/2(n+1)
n=1

Please replies with explanation if possible .......
Hint: A necessary (but not sufficient) condition for the convergence of $\displaystyle \sum_{n = 1}^{\infty} a_n$ is that $\displaystyle \lim_{n \rightarrow \infty} a_n = 0$.

3. ## Please can u evaluate this limit

Evaluate
lim 3n/2(n+1)
n→∞

I have evaluated it but dont know if its right or wrong...

4. Originally Posted by varunnayudu
Evaluate
lim 3n/2(n+1)
n→∞

I have evaluated it but dont know if its right or wrong...
Take out a factor of n on both sides:

$\displaystyle \frac{n}{n} . \frac{3}{2+\frac{1}{n}} = \frac{3}{2+\frac{1}{n}}$

Is that easier?

5. Or you can use l'Hopital's rule.

6. Direct comparison test also works fine:

$\displaystyle \frac{3n}{2n+2}\ge\frac{3n}{2n^{2}+2n^{2}}=\frac{3 n}{4n^{2}}=\frac34\cdot\frac{1}{n}.$

7. Originally Posted by varunnayudu
Test the convergence of the series $\displaystyle \sum_{n=1}^\infty \frac{3n}{2(n+1)}$

Note: Please specify which test to use ,Ratio Test,Integral Test,Raabe Test etc.....
There is probably one that is a little easier to use, but you can use the integral test here:

$\displaystyle \sum_{n=1}^{\infty}\frac{3n}{2\left(n+1\right)}\im plies\tfrac{3}{2}\int_1^{\infty}\frac{x}{x+1}\,dx$

Let $\displaystyle z=x+1\implies x=z-1$. Thus, $\displaystyle \,dz=\,dx$

Thus, the integral becomes $\displaystyle \tfrac{3}{2}\int_2^{\infty}\frac{z-1}{z}\,dz=\tfrac{3}{2}\int_2^{\infty}\left(z-\frac{1}{z}\right)\,dz$

This gives us $\displaystyle \tfrac{3}{2}\left.\left[\tfrac{1}{2}z^2-\ln\left|z\right|\right]\right|_2^\infty=\infty$

Thus, this diverges by the integral test.

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I just thought of an easier way to do this...

You can show that the series is divergent by the nth term test (also called Divergence Test):

$\displaystyle \lim_{n\to\infty}\frac{3n}{2\left(n+1\right)}=\fra c{3}{2}\neq0$. Thus, the series diverges.

8. Originally Posted by Chris L T521
There is probably one that is a little easier to use, but you can use the integral test here:

$\displaystyle \sum_{n=1}^{\infty}\frac{3n}{2\left(n+1\right)}\im plies\tfrac{3}{2}\int_1^{\infty}\frac{x}{x+1}\,dx$

Let $\displaystyle z=x+1\implies x=z-1$. Thus, $\displaystyle \,dz=\,dx$

Thus, the integral becomes $\displaystyle \tfrac{3}{2}\int_2^{\infty}\frac{z-1}{z}\,dz=\tfrac{3}{2}\int_2^{\infty}\left(z-\frac{1}{z}\right)\,dz$

This gives us $\displaystyle \tfrac{3}{2}\left.\left[\tfrac{1}{2}z^2-\ln\left|z\right|\right]\right|_2^\infty=\infty$

Thus, this diverges by the integral test.

[snip]
Integral test requires that $\displaystyle f$ must be positive, continuous and decreasing on $\displaystyle [1,\infty[$ then it applies.

Be sure to verify these conditions before you may apply it.