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Series convergence question

  1. Nov 7, 2005 #1
    I have the following series:
    [tex] S = \sum _{n=0} ^{\infty} 4^n (x+2)^n [/tex]
    Is that the same as [tex]4^n \sum_{n=0} ^{\infty} (x+2)^n = 4^n ((x+2) + (x+3) + \cdots + (x+n))[/tex] ???
    Best Regards Bob
     
    Last edited: Nov 7, 2005
  2. jcsd
  3. Nov 7, 2005 #2
    No... it looks clearly different. What makes you think that?
     
  4. Nov 7, 2005 #3
    Okay thank You I just had to be sure,

    Taking the first sum into account:

    [tex] S = \sum _{n=0} ^{\infty} 4^n (x+2)^n [/tex]

    which method do I use to show for which values of x the series converges?

    Best Regards,

    Bob
     
  5. Nov 7, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Either the ratio test or the root test will work. Typically, the ratio test is easier:

    [tex]\sum_{n=0}^\infty}a_n[/tex], an positive, converges if
    [tex]lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}[/tex] exists and is less than 1.
    Here,
    [tex]\left|a_n\right|= \left|4^n(x+2)^n\right|= \left|(4(x+2))^n\right|[/tex]
    [tex]\left|a_{n+1}\right|= \left|(4(x+2))^{n+1}\right|[/tex]
    so
    [tex]\left|\frac{a_{n+1}}{a_n}\right|= \left|4(x+2)\right|[/tex]

    That will be less than 1 provided |x+2|< 1/4. In other words, for -2-1/4< x< -2+ 1/4 or -9/4< x< -7/4. Of course, you will need to check the endpoints.

    For this example, since we have that "n" power, the ratio test is even easier.
    [tex]\sum_{n=0}^\infty}a_n[/tex], an positive, converges if
    [tex]lim_{n\rightarrow\infty}^n\sqrt{a_n}[/tex] exists and is less than 1.
    [tex]^n\sqrt{\left|(4(x+2)^n\right|}= \left|4(x+2)\right|[/tex]
    so we must, again, have 4|x+2|< 1.

    Actually, it would be much easier to just note that this is a geometric series with common ratio 4(x+ 2)!
     
    Last edited: Nov 7, 2005
  6. Nov 7, 2005 #5
    Hi and thanks You very much for your answer,
    Then the values for which S converges are these:
    -2-1/4< x< -2+ 1/4 or -9/4< x< -7/4 ?

    Or is it the end points?

    Best Regards,
    Bob
     
  7. Nov 7, 2005 #6

    HallsofIvy

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    What I said was "a series converges if the sequence of ratios [itex]\frac{a_{n+1}}{a_n}[/tex] converges to a number less than 1 or if the sequence of roots [itex]^n\sqrt{a_n}[/itex] converges to a number less than 1." What I should have said, but didn't, was that the series diverge if those sequence either diverge or converge to a number larger than 1. If the sequences converge to 1- either can happen. Typically one end point gives an alternating series that converges (by the alternating series test) and the other gives a series that does not converge- although it is possible for the series to converge at both end points. That was shy I said "Of course, you will need to check the endpoints."
     
  8. Nov 8, 2005 #7
    Okay thank you,

    the values for which the series S converges is that then that [tex]x \leq 2 [/tex] cause these are the only values which makes the inequality true?

    Best Regards,

    Bob
     
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