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Question on sequence and limits.

  1. Aug 9, 2006 #1
    prove that the limit of the sequence an=1/n+1/(n+1)+...+1/2n
    exists. show that the limit is less than 1 but not less than 1/2.

    the first part of the question i did already, im not sure about the second part of the question if i did properly.

    because the limit exists, for every n>N(e) (for every e>0) |a-an|<e
    and thus, a-e<an<a+e
    if we let e=1/2 then a+e<=1.5 and thus a<=1.
    thus a>1/2.

    is this correct?

    thanks in advance.
  2. jcsd
  3. Aug 9, 2006 #2


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    I don't follow you.

    You should do this by trying to find sequences whose limits are 1 or 1/2 but that are respectively greater than or less than this sequence. What if every term in the sum were 1/n? What if every term in the sum were 1/2n?
  4. Aug 10, 2006 #3
    thanks for the help, i got it now.
    i have another two questions about limits:
    1)find: lim(1/(1*2*3)+1/(2*3*4)+...+1/n(n+1)(n+2)) as n appraoches infinity.
    2)if a0+a1+...+ap=0, prove that [tex]lim(a_0\sqrt n+a_1\sqrt (n+1)+...a_p\sqrt (n+p))=0[/tex] as n appraoches infinity.
    for the second question i have the hint to take out sqrt(n) out as a factor which i did, but i got this:
    lim(sqrt(n))*lim(a0+a1sqrt(1+1/n)+...+ap(sqrt(1+p/n)) the first limit diverges to infinity, and the second converges to zero, but as i know infinity times 0 is intermediate. (if it were [tex]n^{1/n}[/tex] then it would converge to 1).
    for the first question i tried to use this identity, 1/k(k+1)(k+2)=1/(k+2)[1/k-1/(k+1)] but didnt succeed here.

    thanks in advance.
  5. Aug 10, 2006 #4


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    Use the full partial fraction decomposition:


    here is a tool for partial fractions: www.quickmath.com
  6. Aug 10, 2006 #5


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    I don't know how you were supposed to do the second one, but one way that works:
    The problem is equivalent to the problem of whether [tex]\sqrt{n}-\sqrt{n+1}[/tex] goes to 0. This is because the original sum can be written as a sum of p - 1 constant multiples of these simpler terms. Namely,
    [tex]a_0\sqrt {n}+a_1\sqrt {n+1}+...a_p\sqrt {n+p}[/tex]
    [tex]\sum_{i=0}^{p-1} (\sum_{j=0}^{i}a_j) \cdot (\sqrt{n + i}-\sqrt{n + i+1})[/tex]
    Last edited: Aug 10, 2006
  7. Aug 12, 2006 #6


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    Since [tex]a_0+a_1+\cdots +a_p=0,[/tex] we have [tex]a_0= -\left( a_1+\cdots +a_p\right) ,[/tex] and hence

    [tex] \lim_{n \rightarrow\infty} \left( a_0\sqrt{n}+a_1\sqrt{n+1}+\cdots +a_p\sqrt{n+p}\right) [/tex]
    [tex]= \lim_{n \rightarrow\infty} \left[ -\left( a_1+\cdots +a_p\right) \sqrt{n}+a_1\sqrt{n+1}+\cdots +a_p\sqrt{n+p}\right] [/tex]
    [tex]=\lim_{n \rightarrow\infty} \left[ a_1\left( \sqrt{n+1} - \sqrt{n}\right)+ a_2\left( \sqrt{n+2} - \sqrt{n}\right) + \cdots + a_p\left( \sqrt{n+p} - \sqrt{n}\right) \right] [/tex]
    [tex]=\lim_{n \rightarrow\infty} \left( \frac{a_1}{\sqrt{n+1} + \sqrt{n}}+ \frac{2a_2}{\sqrt{n+2} + \sqrt{n}}+ \cdots + \frac{pa_p}{\sqrt{n+p} + \sqrt{n}} \right) =0 [/tex]
    Last edited: Aug 12, 2006
  8. Aug 12, 2006 #7
    Very nice! :approve:
  9. Aug 12, 2006 #8
    ****, i knew it was simple, anyway here are a couple of more questions:
    1)without reference to binomial theorem show that an=(1+1/n)^n is monotone increasing and bn=(1+1/n)^(n+1) is monotone decreasing.
    (hint: use bernouli inequality).
    well i got these inequalities but i dont know how to infer from them that an+1>=an.
    here are the inequalities (i know you can infer this also from the limit of the sequence):
    what's missing here?
    2) if an>0 and lim a_n+1/a_n=L as n appraoches infinity, then lim (a_n)^(1/n)=L
    what i did is as follows:
    for every n>N(e) (for every e>0) |an+1/an-L|<e, L-e<an+1/an<L+e
    if an+1/an<=1 then the sequence decreasing abnd is bounded below by:
    L-e, cause an+1>an(L-e)>=L-e thus |an^(1/n)-L|<=|an-L|<e.
    if an+1/an>1 then the sequence is increasing and L>=1 thus
    an<an+1/(L-e) is bounded above an+1>an(L-e)>L-e thus
    what do you think of my appraoch? is it valid?
  10. Aug 12, 2006 #9


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    lqg, for the love of god, please use spaces between lines and learn a little latex. It isn't hard, and makes things 100 times easier to read (as well as easier to write).

    Anyway, for the first question, the Bernoulli inequality can be applied if you divide successive terms. For example:

    [tex]\frac{(1+\frac{1}{n})^{n}}{(1+\frac{1}{n-1})^{n-1}}=\left(\frac{ (\frac{n+1}{n}) }{ (\frac{n}{n-1}) }\right)^n (\frac{n}{n-1})=(1-\frac{1}{n^2})^n(\frac{n}{n-1})[/tex]

    Then apply the inequality.
    Last edited: Aug 12, 2006
  11. Aug 12, 2006 #10
    i know little bit latex, but it takes more time to type in latex.
  12. Aug 12, 2006 #11
    You should practice making your proofs more readable, I have added modifications to what you wrote in bold to give you an example of how to clarify your statements.

    This is a false statement, in order for it to be true you must be certain that an >= 1 for all n>N(e) which is not hypothesized. .

    Try this,
    You have that for all n > N,

    [tex] L-e < \left| \frac{a_{n+1}}{a_n} \right| < L + e \ \ \ \ \ \ (1)[/tex]

    for n>N we can write

    [tex] |a_n| = \left| \frac{a_n}{a_{n-1}} \right| \cdot \left| \frac{a_{n-1}}{a_{n-2}} \right| \cdots \left| \frac{a_{N+1}}{a_N} \right| \cdot |a_N| \ \ \ \ \ \ \ \ \ \ \ \ (2)[/tex]

    note that there are n-N fractions so use inequality (1) n-N times.
    Last edited: Aug 12, 2006
  13. Aug 12, 2006 #12
    i dont see how can i use this?
  14. Aug 12, 2006 #13
    Using (1) and (2), we have that

    [tex]|a_N|\cdot(L-e)^{n-N} < |a_n| < |a_N|\cdot(L+e)^{n-N} \ \ \ \ \ \ \ \ \ \ \ \ (3)[/tex]

    define [itex]c_1 = |a_N|\cdot(L-e)^{-N}[/itex], and [itex]c_2 = |a_N|\cdot(L+e)^{-N}[/itex]

    Then (3) becomes:

    [tex]c_1(L-e)^n < |a_n| < c_2(L+e)^n [/tex]

    take the n'th root of each side and then take the limit as n goes to infinity. Recall that if a>0 then [itex] \lim_{n \rightarrow \infty} a^{\frac{1}{n}} = 1[/itex]
    Last edited: Aug 12, 2006
  15. Aug 12, 2006 #14


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    Well, it's roughly the same idea as what I said--a sum of constant multiples of p-1 sequences of the form
    [tex]\sqrt{n + k} - \sqrt{n}[/tex]
  16. Aug 14, 2006 #15
    now about this question i have a follow up question:
    use the above to prove that:
    i tried to rearrange it this way:
    e= lim (1+n)^(1/n)
    but i didnt succeed in using the first question you helped me on status.
    can someone hekp on here?

    thanks in advance.
  17. Aug 14, 2006 #16


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    Isn't this so much nicer:

    That took me less than 5 minutes. Please start writing things this way. You'll get used to it in no time, and it makes our lives much easier. After all, we're helping you, so the least you can do is make it easier for us to do this.

    As far as your question, the last equation above is wrong, and needs to be rewritten using limits. Be careful about mixing up n in the inequality and n in the limit defining e, and you should use two different variables for these.

    The best way to solve this problem is probably to take the logs of each side, and then use an integral to approximate [itex] \ln n! = \sum_k \ln k[/itex].
    Last edited: Aug 14, 2006
  18. Aug 15, 2006 #17
    it's in a chapter before integral is being covered, so im guessing there's a solution without using integrals.
  19. Aug 15, 2006 #18


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    exponential-factorial inequality proof

    this follows from the previous problem: where it was proved that if

    [tex]a_n=\left( 1+\frac{1}{n}\right) ^{n}, \, \, \, b_n=\left( 1+\frac{1}{n}\right) ^{n+1}[/tex]​

    then [tex]a_n[/tex] and [tex]b_n[/tex] are monotone increasing and decreasing sequences, respectfully. Also note that [tex]\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=e[/tex] and so that for every [tex]n\in\mathbb{Z}^{+}[/tex] we have,

    [tex]a_n\leq e \leq b_n[/tex]​

    that is

    [tex]\left( 1+\frac{1}{n}\right) ^{n}\leq e \leq \left( 1+\frac{1}{n}\right) ^{n+1}[/tex]​

    now take the product of each term in this inequality from n=1 to k, i.e.

    [tex]\prod_{n=1}^{k} \left( 1+\frac{1}{n}\right) ^{n}\leq e^{k} \leq \prod_{n=1}^{k}\left( 1+\frac{1}{n}\right) ^{n+1}[/tex]​

    rewrite this as

    [tex]\prod_{n=1}^{k} \left( \frac{n+1}{n}\right) ^{n}\leq e^{k} \leq \prod_{n=1}^{k}\left( \frac{n+1}{n}\right) ^{n+1}[/tex]​

    expand it out

    [tex]\frac{2^1\cdot 3^2\cdot 4^3 \cdots (k+1)^{k}}{1^1\cdot 2^2\cdot 3^3 \cdots k^{k}}\leq e^{k} \leq \frac{2^2\cdot 3^3\cdot 4^4 \cdots (k+1)^{k+1}}{1^2\cdot 2^3\cdot 3^4 \cdots k^{k+1}}[/tex]​

    notice the products telescope, and cancellation gives

    [tex]\frac{(k+1)^{k}}{k!}\leq e^{k} \leq \frac{(k+1)^{k+1}}{k!}[/tex]​

    you can get it from here...
  20. Aug 15, 2006 #19
    after rearranging the last equation, i get:
    if i put k=n-1 i still dont get the desired inequality.
  21. Aug 15, 2006 #20


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    this gives tighter bounds...

    It's there: no worries... in fact, this gives tighter bounds. Here, put k=n-1 to get

    [tex]\frac{n^{n-1}}{e^{n-1}}\leq (n-1)!\leq\frac{n^{n}}{e^{n-1}}[/tex]​

    now multiply by n to get

    [tex]\frac{n^{n}}{e^{n-1}}\leq n!\leq\frac{n^{n+1}}{e^{n-1}}[/tex]​

    rewrite this as

    [tex]e\left( \frac{n}{e}\right) ^{n}\leq n!\leq en\left( \frac{n}{e}\right) ^{n}[/tex]​

    now since [tex]1<e[/tex] and since [tex]n<n+1[/tex] we have

    [tex]\left( \frac{n}{e}\right) ^{n} < e\left( \frac{n}{e}\right) ^{n}\leq n!\leq en\left( \frac{n}{e}\right) ^{n}< e(n+1)\left( \frac{n}{e}\right) ^{n}[/tex]​

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