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Colt1911 [192]
3 years ago
15

Evaluate the limit by first recognizing the sum as the Riemann sum for a function defined on the interval [0,1]: lim as n approa

ches infinity (1/n)((sqrt(1/n)+(sqrt(2/n)+(sqrt(3/n)+...............+(sqrt(n/n))
Mathematics
1 answer:
mina [271]3 years ago
3 0
\lim_{n \to +\infty} \left ( \dfrac{1}{n} \cdot \left ( \sqrt{ \dfrac{1}{n} } + \sqrt{ \dfrac{2}{n} } + \sqrt{ \dfrac{3}{n} } + \cdots + \sqrt{ \dfrac{n}{n} }} \right ) \right ) =

\dfrac{1}{n} \sum_{i = 1}^n \left (  \sqrt{ \dfrac{i}{n} }   \right ) =  \int\limits^1_0 { \sqrt{x} } \, dx  =  \left [  \frac{2}{3}x^{ \frac{3}{2} }       \right ]^{1}_{0} =   \dfrac{2}{3} \cdot 1^{ \frac{3}{2} }  -  \dfrac{2}{3} \cdot 0^{ \frac{3}{2} }  =  \boxed{ \dfrac{2}{3} }
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WILL GIVE BEST RESPONSE
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We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
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</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1, 
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