Page 1 of 2 12 LastLast
Results 1 to 15 of 30

Math Help - Converge or diverge plus show why

  1. #1
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    Converge or diverge plus show why

    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???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Converge or diverge plus show why

    Quote Originally Posted by Shizaru View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    Re: Converge or diverge plus show why

    Quote Originally Posted by Also sprach Zarathustra View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,607
    Thanks
    1574
    Awards
    1

    Re: Converge or diverge plus show why

    Quote Originally Posted by Shizaru View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391

    Re: Converge or diverge plus show why

    Quote Originally Posted by Shizaru View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391

    Re: Converge or diverge plus show why

    Quote Originally Posted by Shizaru View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    Re: Converge or diverge plus show why

    Quote Originally Posted by Prove It View Post
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Converge or diverge plus show why

    Quote Originally Posted by Prove It View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391

    Re: Converge or diverge plus show why

    Quote Originally Posted by alexmahone View Post
    Why is that not sufficient?
    What about the harmonic series? The terms tend to 0 but the series does NOT converge.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391

    Re: Converge or diverge plus show why

    Quote Originally Posted by Shizaru View Post
    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...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    Re: Converge or diverge plus show why

    Quote Originally Posted by alexmahone View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,607
    Thanks
    1574
    Awards
    1

    Re: Converge or diverge plus show why

    Quote Originally Posted by Shizaru View Post
    i mean the 1/5^r being greater than the sin one?
    That is a simple geometric series with ratio less that 1.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391

    Re: Converge or diverge plus show why

    Quote Originally Posted by Shizaru View Post
    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*}
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    Re: Converge or diverge plus show why

    Quote Originally Posted by Prove It View Post
    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?)
    Last edited by Shizaru; January 11th 2012 at 06:42 AM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    Re: Converge or diverge plus show why

    Quote Originally Posted by Plato View Post
    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
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: July 7th 2010, 09:47 PM
  2. Replies: 1
    Last Post: March 29th 2010, 10:27 AM
  3. Converge or diverge
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 15th 2010, 08:58 PM
  4. Converge of Diverge...why?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 8th 2009, 09:31 PM
  5. converge or diverge
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 15th 2009, 03:44 PM

Search Tags


/mathhelpforum @mathhelpforum