Real Analysis: Limit of Sequence

**valerian** · October 19th 2010, 07:14 AM

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.

**emakarov** · October 19th 2010, 09:47 AM

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 $\varepsilon/2$ .

**valerian** · October 19th 2010, 05:26 PM

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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