## anonymous one year ago Series exercie

1. anonymous

$\sum_{1}^{\infty} \frac{ n }{ n ^{2} +1 }$

2. anonymous

@Owlcoffee

3. Owlcoffee

Okay, what's the objective?

4. anonymous

to find if it is convergent,conditonally convergent or divergent

5. anonymous

@ayeshaafzal221 helped me yesterday with it,I just wanna see another way of solving it

6. anonymous

Help?

7. mathmate

Hint: use the integral test.

8. anonymous

I got it like this: $\lim_{n \rightarrow \infty} \frac{ \frac{ n }{ n ^{2} } }{ \frac{ n ^{2} }{ n ^{2} +1}} = \lim_{n \rightarrow \infty} \frac{ n }{ 2 } =\frac{ \infty }{ 2 }=0$

9. anonymous

Is my solution right?

10. ganeshie8

You could also use the comparison test : for $$n\ge 1$$, we have $$n^2+1 \le n^2+n^2 = 2n^2$$ $$\implies \dfrac{1}{n^2+1}\ge \dfrac{1}{2n^2}$$ $$\implies \dfrac{n}{n^2+1}\ge \dfrac{n}{2n^2} = \dfrac{1}{2n}$$ Since $$\sum \dfrac{1}{2n}$$ doesn't converge as this is a harmonic series, the series $$\sum \dfrac{n}{n^2+1}$$ doesn't converge by comparison test

11. anonymous

So my solution is not right?

12. ganeshie8

I would give you 1/10 for the attempt because I don't really get what test you have used..

13. ganeshie8

Also why do you think $$\dfrac{\infty}{2}$$ is anything close to $$0$$ ?

14. mathmate

$$\large\frac{n}{n^2+1}\ne\frac{n/n^2}{n^2/n^2+1}$$ so it is not the equivalent series, in any case $$\inf/2=\inf$$

15. anonymous

i meant infinite,not 0,sorry

16. ganeshie8

do you know what does it mean to 'converge' or 'diverge' ?

17. anonymous

Nope,I know if it dose not equal 0,it means it is divergent,if it is 0,then it's complicated

18. anonymous

does*

19. ganeshie8

before trying the convergence tests, you need to know the meaning of terms "convergent sreies" and "divergent series"

20. anonymous

can I use L'Hospital for this exercise?

21. anonymous

I get 1/2 if I use L'Hospital and it means it is divergent,right?