1. ## Convergence / Divergence

Determine whether the following infinite series are convergent or divergent

i) $\sum_{n=1}^{\infty }{\frac{n}{n+1}}$
ii) $\sum_{n=1}^{\infty }{\frac{n \sqrt{n}}{n^2+1}}$
iii) $\sum_{n=2}^{\infty }{\frac{1}{n^2-1}}$

for part 1 and 2 i tired doing ratio test and integral test but they both failed and i got nowhere. anyone want to give me a hand ?

part 3 seems a lot easier it seems to converge to 0.75 which is easily proved using partial fractions

2. Originally Posted by bobak
for part 1 and 2 i tired doing ratio test and integral test but they both failed and i got nowhere. anyone want to give me a hand ?

part 3 seems a lot easier it seems to converge to 0.75 which is easily proved using partial fractions
for part (i), simply use the test for divergence

for part (ii), still thinking of the nicest way to do this. don't want to try something like the integral test if it's not necessary

for part (iii). rewrite $\frac 1{n^2 - 1}$ as $\frac A{n + 1} + \frac B{n - 1}$ (you can use partial fractions, but you don't have to). this should be a telescopic sum. write out some of the terms so you can see which will cancel out and which will stay. take the limit as $n \to \infty$ when done to see the sum

3. part 1 $\lim_{n \to{\infty}} \frac{n}{n+1} = 1
$
so the series is divergent. that makes perfect sense.

Yeah i got part 3 by the whole method of difference approach i just though i would post it for completeness as it was part of the question

4. Originally Posted by Jhevon
for part (i), simply use the test for divergence

for part (ii), still thinking of the nicest way to do this. don't want to try something like the integral test if it's not necessary

for part (iii). rewrite $\frac 1{n^2 - 1}$ as $\frac A{n + 1} + \frac B{n - 1}$ (you can use partial fractions, but you don't have to). this should be a telescopic sum. write out some of the terms so you can see which will cancel out and which will stay. take the limit as $n \to \infty$ when done to see the sum
(ii) Note that $\frac{n \, \sqrt{n}}{n^2 + 1} \geq \frac{n \, \sqrt{n}}{n^2 + n^2} = \frac{n \, \sqrt{n}}{2n^2} = \frac{1}{2} \, \frac{1}{\sqrt{n}}$, for $n \geq 1$.

And you know that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ is divergent.

From the comparison test, the divergence of $\sum_{n=1}^{\infty} \frac{n \, \sqrt{n}}{n^2 + 1}$ is easily seen.

5. Originally Posted by mr fantastic
(ii) Note that [tex]

And you know that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ is divergent.
How do we know this ?

6. Originally Posted by bobak
How do we know this ?
Because,
$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1} {\sqrt{n}} \geq \int_1^{n+1} \frac{dx}{\sqrt{x}} = 2\sqrt{n+1} - 2$.

7. Originally Posted by bobak
How do we know this ?
it is a divergent p-series

$\sum \frac 1{n^p}$ converges if $p>1$, diverges if $p \le 1$

TPH showed you a more rigorous/specific proof, but this is what it summarizes to in general