Question

double check convergence of series

Answer 1

\[\sum_{n=1}^{\infty} (-1)^{n} \sqrt{\frac{ n+1 }{ 4n+1 }}\]

Answer 2

I said conditionally convergent because the limit of a_n is 1/2 so it would be divergent for the absolute value but with the -1 it would converge because of the alternating series

Answer 3

seems to be inconclusive for root and ratio tests

Answer 4

lim a_n must be zero for a series to converge at all, this includes alternating series

Answer 5

since the limit of a_n is not zero, your series diverges

Answer 6

ok

Answer 7

thank you I needed that clarified

Answer 8

$$ \Large
{
\text{Given } \sum a_n\\
if~~ lim_{n \to \infty } |a_n| \neq 0~~ then \sum a_n ~ diverges

}
$$

Answer 9

the sequence diverges as well.

Answer 10

@SolomonZelman the positive values of the sequence converge to 1/2
and are you sure that it diverges with (-1)^n .

Answer 11

the denominator is getting larger so it is decreasing so it would be conditionally convergent then?

Answer 12

The more general theorem is:
$$
\Large
{
\text{Given } \sum a_n\\
if~ \lim_{n \to \infty } a_n \neq 0~~ then \sum a_n ~ diverges

}

$$

Answer 13

so it is totally divergent then

Answer 14

yes

Answer 15

$$
\Large
{
\text{Given } \sum a_n\\
if~~ \lim_{n \to \infty } a_n \neq 0~~ then \sum a_n ~ diverges \\~\\

\text{Given} ~\sum (-1)^n \cdot b_n\\
if~~ \lim_{n \to \infty }~ b_n \neq 0~~ then \sum (-1)^n \cdot b_n ~ diverges

}

$$

Answer 16

but you can probably combine these two results into one

Answer 17

oh I see, the original sequence does diverge.
$$
\Large {

\lim_{n \to \infty} (-1)^n\cdot \frac{\sqrt{n+1} }{ \sqrt{4n+1} }
= \begin{cases}
1/2 ~if ~n ~is ~even
\\-1/2 ~ if~n ~is~ odd
\end{cases}

}
$$

Answer 18

if the limit of A_n as n -> \(\large\color{slate}{\displaystyle\infty}\) alternates, than sequence diverges, because the limit doesn't exist.

(the terms in the sequence don't approach any particular value)

Answer 19

wolfram gives a strange answer
http://www.wolframalpha.com/input/?i=lim%28%28-1%29^n*sqrt%28n%2B1%29%2Fsqrt%284*n%2B1%29%2C+n+%3D+infinity%29

Answer 20

I am pretty sure that if the limit doesn't exist (alternates) the sequence is divergent.

Answer 21

Not that I ever claimed to be anywhere close to a mathematician or anything of this sort, but I know that is

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}A_n}\) doesn't exist - in this case alternates, the sequence doesn't converge.

For sequence to converge, the terms in the sequence (for any INFINITYth term, roughly speaking) should be approaching ONE SPECIFIC value.

like for example,
{-3, 2, -3, 2, -3, 2 ...} is a divergent sequence
or {-6,6,-6,6,-6,-6} is also divergent

Answer 22

anyways, sequence converges wasn't the initial discussion, just posting for the interest.

Answer 23

right.
if the limit of a_n does not exist, then how much more so it does not equal to zero.

Answer 24

$$
\Large
{
\text{Given } \sum a_n\\
if~~ \lim_{n \to \infty } a_n \neq 0~~ then \sum a_n ~ diverges \\~\\

}
$$

Answer 25

if lim a_n does not exist, then lim a_n does not equal to zero, then series a_n diverges

Answer 26

this might be called an "a fortiori argument"