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Homework Help: Help with series problem

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Problem is attached in this post.

    2. Relevant equations

    Problem is attached in this post.

    3. The attempt at a solution

    I don't know how to approach this problem at all. How exactly do I find what function an is? Also what do I do after finding an?

    What I did was let an=n then ended up doing Lim 1/n as n -> ∞ and ended up getting 0 as my answer.

    The correct answer is that the series diverges.

    Attached Files:

  2. jcsd
  3. Apr 2, 2014 #2
    You do not need to find an, as there are in fact many different series that converge to e. Instead, consider the necessary criteria for a series to converge. What behavior is required of the terms an in order for the series to converge to any value ?
  4. Apr 2, 2014 #3


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    Once you realise that ##\lim_{n \to \infty} a_n = 0##, you already have an answer. What is ##\lim_{n \to \infty} \frac{1}{a_n}##?
  5. Apr 2, 2014 #4
    The only behavior that I'm aware of is that if an is convergent then, Lim an n -> ∞ = 0.

    However I still don't understand how I can use that specific behavior of an to solve this problem?
  6. Apr 2, 2014 #5
    That's exactly what I'm confused about, does an = 0, since its Limit at ∞ is 0? What I mean is, do I substitute an as equal to 0?
  7. Apr 2, 2014 #6


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    This is very strange. You seem to be confused about the converence of series and sequences in general.
    Look at an example. You know that [itex]\lim_{n\to\infty} 1/n= 0[/itex] don't you? And that 1/n is never equal to 0?

    So if [itex]a_n= 1/n[/itex], what can you say about [itex]1/a_n[/itex]?
  8. Apr 2, 2014 #7


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    ##a_n## isn't 0. But it becomes very small as n->infinity. So?
  9. Apr 2, 2014 #8
    Excellent! If [tex]\lim_{n\rightarrow \infty} a_n = 0,[/tex] what can we say about [tex]\lim_{n\rightarrow\infty} \frac{1}{a_n}\; ?[/tex] If you are not sure, think about the epsilon-delta definition of the first limit, and consider what that implies for the term 1/an.
    We cannot identify the value of a function with its limit! For example, [tex]\lim_{n\rightarrow\infty} \frac{1}{n} = 0,[/tex] but there is no particular value of n for which 1/n = 0.
  10. Apr 2, 2014 #9
    Would that be enough to then say that 1/an diverges?
  11. Apr 2, 2014 #10
    I haven't actually provided a coherent argument, just a question. If you answer the question in the right way, it would be enough. That is, if you can prove that the limit above is not 0, then you may use the contrapositive of your criterion (which is logically equivalent) to imply that the associated series of terms diverges.
    To refresh, your criterion stated that "If the series [itex]\sum a_n[/itex] converges, then [itex]\lim_{n\rightarrow\infty} a_n = 0[/itex]". The contrapositive statement is equivalent and is therefore always implied: "If [itex]\lim_{n\rightarrow\infty} a_n \neq 0[/itex], then the series [itex]\sum a_n[/itex] does not converge."
    So we only need to show that the limit is not 0. We do not need to know what the limit is, or even if it exists. However, if you know this limit's value, and it is not 0, you may use that information as well.
    If you are uncertain when dealing with limits, you will need to go back to the definition. The statement "[itex]\lim_{n\rightarrow\infty} \frac{1}{a_n} = 0[/itex]", which we actually want to be false, is defined by the criteria that "For all values of ε > 0, and for all natural numbers n, there is some number N such that, if n > N, then [itex]\left|\frac{1}{a_n} - 0\right| < \epsilon[/itex]." Intuitively, it says "eventually, the distance between 1/an and 0 becomes less than any positive number."
    We want to show that this limit is not 0, and thus we want to show that the above statement is false. One way to do this is to prove the logical negation of this statement: "There exists some value of ε > 0 such that, for all numbers N, there is some natural number n > N such that [itex]\left|\frac{1}{a_n} - 0\right| \geq \epsilon[/itex]".
    That is, we want to show that the distance between [itex]\frac{1}{a_n}[/itex] and 0 can never be made smaller than a particular amount ε. The key to finding this particular amount is to use the fact that [itex]\lim_{n\rightarrow\infty} a_n = 0[/itex]. That is, we need the epsilon-N statement that defines this limit and a tiny bit of algebra to show us exactly what [itex]\left|\frac{1}{a_n}\right|[/itex] is always greater than (at least once we get to high enough values of n, which is what N is for).
    Last edited: Apr 2, 2014
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