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Math Help - Real Analysis: Limit of Sequence

  1. #1
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    Real Analysis: Limit of Sequence

    There exists a sequence (x_{n}) such that \lim_{n \longrightarrow \infty} x_{n} = x and there exists a sequence (y_{n}) of non-negative real numbers such that \lim_{n \longrightarrow \infty} y_{1} + y_{2} + y_{3} + ... + y_{n-1} + y_{n}  = \infty .

    Show:  \lim_{n \longrightarrow \infty} \frac{x_{1}y_{1} + x_{2}y_{2}  + ... + x_{n-1}y_{n-1} + x_{n}y_{n}}{y_{1} + y_{2} + ... + y_{n-1} + y_{n}}  = x .

    I'm having some trouble figuring out where to begin on this one. I've tried re-writing it in summation notation and manipulating that in hopes of being able to cancel y terms but hasn't gotten me anywhere. I thought it might be possible to rewrite the expression as a subsequence of (x_{n}) and then take the fact that a subsequence converges to the same limit as the original sequence but I would not no where to begin to construct the subsequence.
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  2. #2
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    I've tried re-writing it in summation notation and manipulating that in hopes of being able to cancel y terms but hasn't gotten me anywhere.
    If algebraic manipulations fail, I find it useful to think intuitively about what happens at large n. The idea is that x_i will be become sufficiently close to x, so, roughly speaking, they can be factored out.

    Let \Delta x_i=x_i-x. Then \displaystyle\frac{x_{1}y_{1} + x_{2}y_{2} + ... + x_{n-1}y_{n-1} + x_{n}y_{n}}{y_{1} + y_{2} + ... + y_{n-1} + y_{n}}-x=\frac{\Sigma_{i=1}^n\Delta x_iy_i}{\Sigma_{i=1}^n y_i}. Suppose \varepsilon is given and we need to make this fraction less than \varepsilon (in absolute value) by choosing a sufficiently large n.

    We represent the fraction as a sum \displaystyle\frac{\Sigma_{i=1}^m\Delta x_iy_i}{\Sigma_{i=1}^n y_i}+\frac{\Sigma_{i=m+1}^n\Delta x_iy_i}{\Sigma_{i=1}^n y_i} where m is chosen so that \Delta x_i<\varepsilon/2 for i>m. Then the second term, regardless of n, is less than \varepsilon/2. For this given m, the numerator of the first term is fixed, and since the denominator diverges, one can choose n so that the first term is also less than \varepsilon/2.
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  3. #3
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    Thank you. This makes a lot of sense. I tried using an epsilon argument but I didn't think to consider factoring out the x terms as they became extremely close to the limit.
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