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Math Help - sequence, convergence

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    sequence, convergence

    Suppose that (x_n) is a sequence in \mathbb{R}. Define a sequence (y_n) by y_n=\frac{x_1+x_2+\cdots+x_n}{n},  \forall n \in \mathbb{N}. Prove that if (x_n) converges to x \in \mathbb{R} then (y_n) converges to x.
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    Hello,
    Quote Originally Posted by xboxlive89128 View Post
    Suppose that (x_n) is a sequence in \mathbb{R}. Define a sequence (y_n) by y_n=\frac{x_1+x_2+\cdots+x_n}{n},  \forall n \in \mathbb{N}. Prove that if (x_n) converges to x \in \mathbb{R} then (y_n) converges to x.
    This is Cesaro's mean. And you want to prove Cesaro's lemma.
    There is a proof in the French Wikipedia (Lemme de Cesŕro - Wikipédia), but not in the English one.
    I'll translate and adapt it here...


    x_n \to x can be translated this way :
    \forall \varepsilon >0,~ \exists N \in \mathbb{N},~ \forall n>N,~ |x_n-x|< \varepsilon/2

    Let S_N=\sum_{k=1}^N |x_k-x|

    Let's assume n>N :
    |y_n-x|=\left|\frac 1n \sum_{k=1}^n (x_k-x)\right|
    By the triangle inequality, we have :
    |y_n-x| \leqslant \frac 1n \sum_{k=1}^N |x_k-x|+\frac 1n \sum_{k=N+1}^n |x_k-x|
    |y_n-x| \leqslant \frac{S_N}{n}+\frac{n-N}{n} (\varepsilon/2)=\frac{S_N}{n}+\varepsilon/2 -\frac{N}{n} (\varepsilon/2) \leqslant \frac{S_N}{n}+\varepsilon/2
    (because N/n (epsilon/2) >0)

    But S_N doesn't depend on n. So \lim_{n \to \infty} \frac{S_N}{n}=0
    This means that for any \varepsilon >0, there exists an integer N' such that for all n>N', we have :
    \left|\frac{S_N}{n}\right|\leqslant \varepsilon/2


    By combining this latter inequality to the previous one, we have, for n>\max\{N,N'\} :
    |y_n-x| \leqslant \varepsilon/2+\varepsilon/2=\varepsilon \qquad \square
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