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Math Help - A little help with a limit of a series, actuallly two...

  1. #1
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    A little help with a limit of a series, actuallly two...

    could someone help me with the attached problems? Most appreciated.
    Attached Thumbnails Attached Thumbnails A little help with a limit of a series, actuallly two...-limit-q-1.jpg  
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  2. #2
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    Quote Originally Posted by azarue View Post
    could someone help me with the attached problems? Most appreciated.
    The first limit is certainly 0, for consider:

    0\leq \frac{n}{\sum_{k=1}^n\frac{k}{|\sin(k)|}} <br />
\leq \frac{n}{\sum_{k=1}^n k}=\frac{n}{\frac{n(n+1)}{2}}
    =\frac{2n}{n^2+n}=\frac{2}{n+\frac{1}{n}}\rightarr  ow 0,\text{ for }n\rightarrow \infty
    the second inequality sign holds true, because from |\sin(k)|\leq 1 it follows that \frac{k}{|\sin(k)|}\geq k, and thus
    \sum_{k=1}^n \frac{k}{|\sin(k)|}\geq \sum_{k=1}^n k.
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  3. #3
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    Quote Originally Posted by azarue View Post
    could someone help me with the attached problems? Most appreciated.


    Since 0<|\sin n|<1\,\,\,\forall\,n\in\mathbb{N} (why? Hint: \pi is irrational), we get that \frac{n}{|\sin n|}>n\Longrightarrow \frac{n}{|\sin n|}\xrightarrow [n\to\infty]{}\infty \Longrightarrow \frac{\sum^n_{i=1}\frac{i}{|\sin i|}}{n}\xrightarrow [n\to\infty]{}\infty as well, and from here

    that your limit is...


    As for the second one: \left(1-\frac{2}{n+2}\right)^{n+2}\xrightarrow [n\to\infty]{}e^{-2}\Longrightarrow the limit is...

    Theorems used: if 0\leq a_n\xrightarrow [n\to\infty]{}a then \frac{a_1+a_2+\ldots+a_n}{n}\xrightarrow [n\to\infty]{}a and also \sqrt[n]{a_1\cdot a_2\cdot\ldots\cdot a_n}\xrightarrow [n\to\infty]{}a , with the obvious generalizations when a=\infty

    Tonio
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    Thank you for your post.

    tonio,

    in the original question we had to compute a limit who looked something like this: (n)/(sum of something). in your solution you took the opposite, (some of something/n).
    <br />
\Longrightarrow \frac{\sum^n_{i=1}\frac{i}{|\sin i|}}{n}\xrightarrow [n\to\infty]{}\infty<br />

    Could you clarify please ?

    In addition, may I ask for more detailed solution for question number 2.

    Thank you for the help.
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  5. #5
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    Quote Originally Posted by azarue View Post
    tonio,

    in the original question we had to compute a limit who looked something like this: (n)/(sum of something). in your solution you took the opposite, (some of something/n).
    <br />
\Longrightarrow \frac{\sum^n_{i=1}\frac{i}{|\sin i|}}{n}\xrightarrow [n\to\infty]{}\infty<br />

    Could you clarify please ?


    This is the generalization I talked about in my post: you figure this (easy) one out and deduce your limit is zero


    In addition, may I ask for more detailed solution for question number 2.


    Look at the geometric mean sequence that follows from the one in my post. You can also try to google "arithmetic and geometric means limits".
    For example, look at Limit of arithmetic mean , or here More Challenging Limits ( the second one contains proofs of both theorems I quote)

    Tonio

    Thank you for the help.
    .
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  6. #6
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    limits studies

    Tonio, I spent some time reading books on the subject. thank you for you help.
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