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Limit and Series

  1. Dec 2, 2006 #1
    Ok, so I have the problem lim n-> infinity (1 +5/n)^(4n)

    So looking at it without trying anything I can see for n arbitrarily large 5/n goes to 0. That means (1+0)^(infinity). One to the power of any real number is one. However by looking at the definition of e as x->infinity I can say (1+ 1/x)^x=e^1

    So (1+5/x)^(4x)= e^5*4 or e^20, which is not quite one. Where am I going wrong here?

    Also I have the problem lim n-> infinity (n/n+3)^2n . I can same the same thing here as n tends to go to infinity I would have infinity over infinity which is 1? so one raised to infinity is one. I know I am incorrect in this assumption but why?

    How would I go on solving this problem?

    LIM N-> infinity
    (N/N+3)^(2N) =
  2. jcsd
  3. Dec 2, 2006 #2
    [tex] e^{x} = \lim_{n\rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}[/tex]

    [tex] \lim_{n\rightarrow \infty} (1+\frac{5}{n})^{4n} = \left[\left(1+\frac{5}{n}\right)^{n}\right]^{4} = e^{20} [/tex]

    [tex] \left(\frac{n}{n+3}\right)^{2n} = \left[\left(\frac{n}{n+3}\right)^{n}\right]^{2} [/tex]
    Last edited: Dec 2, 2006
  4. Dec 2, 2006 #3


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    Note that:
    [tex](\frac{n}{n+3})^{2n}=(1+\frac{-3}{n+3})^{2(n+3)-6}}=(1+\frac{-3}{n+3})^{-6}*(1+\frac{-3}{n+3})^{2(n+3)}=K_{n}*((1+\frac{-3}{u})^{u})^{2}, K_{n}=(1+\frac{-3}{n+3})^{-6}, u=n+3[/tex]
    Last edited: Dec 2, 2006
  5. Dec 2, 2006 #4


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    "infinity" is not a real number so "One to the power of any real number is one" is irrelevant in talking about "[itex]1^\infty[/itex]" which is really just shorthand for the limit as the exponent gets larger without bound.

    "Here" you are not doing anything wrong, except leaving out the "limit" part.
    (1+ 5/x)^(4x)= ((1+ 5/x)^x)^4. If you let y= x/5, then x= 5y so (1+ 5/x)^x= (1+ 1/y)^(5y)= ((1+ 1/y)^y)^5. That is, for all x, (1+ 5/x)^(4x)= (((1+ 1/y)^y)^5)^4. Clearly, y goes to infinity as x does (more precisely y increases without bound as x does) so the limit (1+ 5/x)^(4x)= lim(((1+ 1/y)^y)^5)^4= e^20 just as you say. There is no reason to think that it should be 1.

    "Infinity over infinity" is NOT 1! Just as "infinity" is not a number so "infinity over infinity" is not a number. The limit depends upon HOW the numerator and denominator go to infinity. For example, 5n/n, as n goes to infinity gives the form "infinity over infinity" but the limit is clearly 5. For a any real number, an/n gives "infinity over infinity" but the limit is clearly a.

    I would rewrite it as [itex]\left(\frac{n}{n+3}\right)^{2n}= \left(\frac{n+3}{n}\right)^{-2n}[/itex][itex]= \left(\left(1+ \frac{3}{n}\right)^n\right)^{-2}[/itex]
    Now that's exactly like the one before.
    Last edited by a moderator: Dec 2, 2006
  6. Dec 2, 2006 #5
    I'll try to explain in two parts:

    PART 1:

    When you examined the sequence [itex]c_{n}=1+ \frac{5}{n}[/itex] you correctly calculated the limit as 1 by doing the following:

    [tex]\lim c_{n} = \lim 1 + \lim \frac{5}{n} = 1+0 = 1[/tex]

    Here, you correctly applied theorem 3 from
    http://imai.princeton.edu/teaching/files/sequences.pdf [Broken]

    [tex]\lim a_{n}+\lim b_{n}=a+b[/tex] with [tex]\lim a_{n}=a[/tex] and [tex]\lim b_{n}=b[/tex]

    PART 2:

    Now comes where you went wrong. Your sequence is

    [tex]d_{n}= \left(1+ \frac{5}{n} \right)^{4n}=\left(c_{n} \right)^{4n}[/tex]

    You then calculated the limit the following (wrong) way:

    [tex]\lim d_{n}=\lim \left(c_{n} \right) ^{4n} =\lim \left( \lim c_{n} \right)^{4n}= \lim \left(1 \right) ^{4n} = 1[/tex]

    The mistake is that you have assumed that

    [tex]\lim \left(a_{n} \right)^{b_{n}} = \lim a^{b_{n}}[/tex] with [tex]\lim a_{n}=a[/tex]

    is a theorem, which is obviously not.

    Thus, unlike in PART 1 where you correctly used a theorem you didn't use a theorem in PART 2.
    Last edited by a moderator: May 2, 2017
  7. Dec 2, 2006 #6
    Thanks for the reply's I got it, for the 2nd problem I get e^-6 which is believe is right.

    Great replies ty.
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