Converge or diverge plus show why

**Shizaru** · January 9th 2012, 07:41 AM

show why the con/diverge:

$\sum_{r=1}^{\infty}\frac{r^{2}-1}{r^{2}+1}$

i think she diverges. because the terms get bigger (closer to 1 each) so you are just adding almost 1 + 1 +1 +1 ... it tails off to infinity... diverge... but i think i need to use a rigurous proof not just intuition.

$\sum_{r=1}^{\infty}\frac{\sin(r^{2})}{5^{r}}$

i'd say she converges since the denom gets rapidly massive so the whole thing is getting rapidly tiny... i can't see what it converges too though? and how to show it?

$\sum_{r=1}^{\infty}\frac{(3r)!}{(r!)^{3}}$

as above, i think its the same.

but i don't know how to show these using the whole episolon thing. the epsilon thing really bugs me. do i need the epsilon thing? can someone help me to show these things rather than just use intuition... also am i right or wrong???

**Also sprach Zarathustra** · January 9th 2012, 08:12 AM

Originally Posted by Shizaru

show why the con/diverge:

$\sum_{r=1}^{\infty}\frac{r^{2}-1}{r^{2}+1}$

i think she diverges. because the terms get bigger (closer to 1 each) so you are just adding almost 1 + 1 +1 +1 ... it tails off to infinity... diverge... but i think i need to use a rigurous proof not just intuition.

$\sum_{r=1}^{\infty}\frac{\sin(r^{2})}{5^{r}}$

i'd say she converges since the denom gets rapidly massive so the whole thing is getting rapidly tiny... i can't see what it converges too though? and how to show it?

$\sum_{r=1}^{\infty}\frac{(3r)!}{(r!)^{3}}$

as above, i think its the same.

but i don't know how to show these using the whole episolon thing. the epsilon thing really bugs me. do i need the epsilon thing? can someone help me to show these things rather than just use intuition... also am i right or wrong???

In the first one $\lim_{r\to\infty} a_r=1$ , so... ?

The second one: $|\frac{\sin(r^{2})}{5^{r}}|<\frac{1}{|5^{r}|}$

In the third: use Ratio test - Wikipedia, the free encyclopedia

**Shizaru** · January 9th 2012, 08:58 AM

Originally Posted by Also sprach Zarathustra

In the first one $\lim_{r\to\infty} a_r=1$ , so... ?

The second one: $|\frac{\sin(r^{2})}{5^{r}}|<\frac{1}{|5^{r}|}$

In the third: use Ratio test - Wikipedia, the free encyclopedia

1st - divergent by nonnull tesT??

2nd - i still dont see enough info, i would try the comparison test, again i come up short? or is this better to use absolute convergence properties?

**Plato** · January 9th 2012, 01:12 PM

Originally Posted by Shizaru

2nd - i still dont see enough info, i would try the comparison test, again i come up short? or is this better to use absolute convergence properties?

The point is if a series converges absolutely it converges period.

**Prove It** · January 9th 2012, 03:24 PM

Originally Posted by Shizaru

show why the con/diverge:

$\sum_{r=1}^{\infty}\frac{r^{2}-1}{r^{2}+1}$

i think she diverges. because the terms get bigger (closer to 1 each) so you are just adding almost 1 + 1 +1 +1 ... it tails off to infinity... diverge... but i think i need to use a rigurous proof not just intuition.

$\sum_{r=1}^{\infty}\frac{\sin(r^{2})}{5^{r}}$

i'd say she converges since the denom gets rapidly massive so the whole thing is getting rapidly tiny... i can't see what it converges too though? and how to show it?

$\sum_{r=1}^{\infty}\frac{(3r)!}{(r!)^{3}}$

as above, i think its the same.

but i don't know how to show these using the whole episolon thing. the epsilon thing really bugs me. do i need the epsilon thing? can someone help me to show these things rather than just use intuition... also am i right or wrong???

A necessary (but not sufficient) condition for a series to converge is that the individual terms have to tend to 0.

Therefore, a valid way to show that a series diverges is to show that the individual terms do NOT tend to 0.

For the first

$\displaystyle \begin{align*} \lim_{r \to \infty}\frac{r^2 - 1}{r^2 + 1} &= \lim_{r \to \infty}\frac{r^2 + 1 - 2}{r^2 + 1} \\ &= \lim_{r \to \infty}1 - \frac{2}{r^2 + 1} \\ &= 1 - 0 \\ &= 1 \end{align*}$

Clearly, the terms do not tend to 0, so the series diverges.

**Prove It** · January 9th 2012, 03:31 PM

Originally Posted by Shizaru

1st - divergent by nonnull tesT??

2nd - i still dont see enough info, i would try the comparison test, again i come up short? or is this better to use absolute convergence properties?

Think about it like this. Suppose you have some series. Since there may be some negative values in it, the sum will never be any greater than the sum of the absolute values of the terms (since they are all positive). Therefore, by the comparison test, if the "larger series" (the series of absolute values) converges, then so must the "smaller series" (the original series).

So for your second series, by showing that $\displaystyle \begin{align*} \sum{\left| \frac{ \sin{\left(r^2\right)} }{ 5^r } \right|} \end{align*}$ converges, you show $\displaystyle \begin{align*} \sum{\frac{\sin{\left(r^2\right)}}{5^r}} \end{align*}$ also converges.

**Shizaru** · January 9th 2012, 03:41 PM

Originally Posted by Prove It

A necessary (but not sufficient) condition for a series to converge is that the individual terms have to tend to 0.

Therefore, a valid way to show that a series diverges is to show that the individual terms do NOT tend to 0.

For the first

$\displaystyle \begin{align*} \lim_{r \to \infty}\frac{r^2 - 1}{r^2 + 1} &= \lim_{r \to \infty}\frac{r^2 + 1 - 2}{r^2 + 1} \\ &= \lim_{r \to \infty}1 - \frac{2}{r^2 + 1} \\ &= 1 - 0 \\ &= 1 \end{align*}$

Clearly, the terms do not tend to 0, so the series diverges.

thats what we call nonnull test. so am i right there (do you think i would need to prove the sequence of terms converges to 1 or can i just state it doesnt converge to 0 by intuition?)

as for the 2nd one... ok i am using the comparison test, and the property of absolute convergence... yes this is one of the properties we are told. i think i get that one now but how do you know the example you gave is always greater (or equal) to the series in question?

my main problem is knowing what needs to be shown and what can just be stated... :S
see my intuition was correct
but i dont always know how to show it

**alexmahone** · January 9th 2012, 03:58 PM

Originally Posted by Prove It

A necessary (but not sufficient) condition for a series to converge is that the individual terms have to tend to 0.

Why is that not sufficient?

**Prove It** · January 9th 2012, 04:14 PM

Originally Posted by alexmahone

Why is that not sufficient?

What about the harmonic series? The terms tend to 0 but the series does NOT converge.

**Prove It** · January 9th 2012, 04:18 PM

Originally Posted by Shizaru

how do you know the example you gave is always greater (or equal) to the series in question?

Because each term can never be any greater than its absolute value...

**Shizaru** · January 10th 2012, 02:49 AM

Originally Posted by alexmahone

Why is that not sufficient?

its not sufficient because it doesnt hold for vice versa. they can converge to 0 without the series converging.

Prove It - i mean the 1/5^r being greater than the sin one?

**Plato** · January 10th 2012, 03:01 AM

Originally Posted by Shizaru

i mean the 1/5^r being greater than the sin one?

That is a simple geometric series with ratio less that 1.

**Prove It** · January 11th 2012, 04:16 AM

Originally Posted by Shizaru

its not sufficient because it doesnt hold for vice versa. they can converge to 0 without the series converging.

Prove It - i mean the 1/5^r being greater than the sin one?

You should know that $\displaystyle \begin{align*} |\sin{X}| \leq 1 \end{align*}$ for all $\displaystyle \begin{align*} X \end{align*}$ .

Therefore

$\displaystyle \begin{align*} \left| \sin{ \left( r^2 \right) } \right| &\leq 1 \\ \frac{ \left| \sin{ \left( r^2 \right) } \right|}{ \left| 5^r \right| } &\leq \frac{1}{ \left| 5^r \right| } \\ \left| \frac{\sin{\left(r^2\right)}}{5^r} \right| &\leq \left| \frac{1}{5^r} \right| \end{align*}$

**Shizaru** · January 11th 2012, 06:17 AM

Originally Posted by Prove It

You should know that $\displaystyle \begin{align*} |\sin{X}| \leq 1 \end{align*}$ for all $\displaystyle \begin{align*} X \end{align*}$ .

of course i know this and i am an idiot for not seeing this was neccessary. thanks for the pointer

but wait doesnt $1 \geq |\sin x|$ imply $1 \geq \sin x$ anyway so why do i have to bother with using absolute convergence properties in the first place? can't i just directly use the comparison with the |1/5^r|?

also we would need to show that 1/5^r converges, but how?

(can someone just confirm my suspicion here also - that it is sinr^2 and not just sin r, actually makes no difference here? is this just an attempt to deceive?)

**Shizaru** · January 11th 2012, 07:00 AM

Originally Posted by Plato

That is a simple geometric series with ratio less that 1.

how is it? it's not (1/5)^r. If it was I could see that

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