Sequence...

Dec 2009
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Russia
Let \(\displaystyle (a_n)^\infty_{n=1}\) converges to \(\displaystyle a\).

Is the next sequence \(\displaystyle b_n\) also converges?

\(\displaystyle b_n=\frac{a_1^2+a_2^2+\dots +a_n^2}{n}\)


Thanks!
 

Plato

MHF Helper
Aug 2006
22,507
8,664
Let \(\displaystyle (a_n)^\infty_{n=1}\) converges to \(\displaystyle a\).
Is the next sequence \(\displaystyle b_n\) also converges?
\(\displaystyle b_n=\frac{a_1^2+a_2^2+\dots +a_n^2}{n}\)
Do you know the idea of the sequence of means?
Because \(\displaystyle \left(a_n\right)\) converges it is true that \(\displaystyle \left(a_n^2\right)\) also converges.