# Thread: Real Analysis: Limit of Sequence

1. ## Real Analysis: Limit of Sequence

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

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

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 $\displaystyle (x_{n})$ 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.

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

We represent the fraction as a sum $\displaystyle \displaystyle\frac{\Sigma_{i=1}^m\Delta x_iy_i}{\Sigma_{i=1}^n y_i}+\frac{\Sigma_{i=m+1}^n\Delta x_iy_i}{\Sigma_{i=1}^n y_i}$ where $\displaystyle m$ is chosen so that $\displaystyle \Delta x_i<\varepsilon/2$ for $\displaystyle i>m$. Then the second term, regardless of $\displaystyle n$, is less than $\displaystyle \varepsilon/2$. For this given $\displaystyle m$, the numerator of the first term is fixed, and since the denominator diverges, one can choose $\displaystyle n$ so that the first term is also less than $\displaystyle \varepsilon/2$.

3. 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.