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Post Posted: Mar 02, 2008, 7:54 am • # 1 


Find \lim_{n\to\infty}a_n where (a_n)_{n\ge1} is defined by a_n=\frac1{\sqrt{n^2+8n-1}}+\frac1{\sqrt{n^2+16n-1}}+\frac1{\sqrt{n^2+24n-1}}+\ldots+\frac1{\sqrt{9n^2-1}}.

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Post Posted: Mar 03, 2008, 12:26 pm • # 2 


I=\lim_{n\to \infty}{a_n}
I=\lim_{n\to \infty}{\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+8kn-1}}}}=\lim_{n\to \infty}{\frac{1}{n}\sum_{k=1}^{n}{\frac{1}{\sqrt{...
 
 
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