Question

Series exercie

Answer 1

\[\sum_{1}^{\infty} \frac{ n }{ n ^{2} +1 }\]

Answer 2

@Owlcoffee

Answer 3

Okay, what's the objective?

Answer 4

to find if it is convergent,conditonally convergent or divergent

Answer 5

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

Answer 6

Help?

Answer 7

Hint: use the integral test.

Answer 8

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

Answer 9

Is my solution right?

Answer 10

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

Answer 11

So my solution is not right?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

i meant infinite,not 0,sorry

Answer 16

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

Answer 17

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

Answer 18

does*

Answer 19

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

Answer 20

can I use L'Hospital for this exercise?

Answer 21

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