Find \(\displaystyle\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{{n}}}}{\frac{{{2}{n}}}{{{\left({n}+{2}{i}\right)}^{{2}}}}}\)

palmantkf4u 2022-04-05 Answered
Find
limni=1n2n(n+2i)2
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Answers (2)

Drahthaare89c
Answered 2022-04-06 Author has 19 answers
The expression
i=1n2n(n+2i)2=1ni=1n2(1+2in)2
is the Riemann sum of the function f(x)=2(1+2x)2 over the interval I=[0,1], corresponding to the uniform partition of I into n equal parts. Since f is Riemann-integrable (being a continuous function over a closed and bounded interval), this sum approaches the integral of f over I as n. That is,
limn1ni=1n2(1+2in)2=012(1+2x)2dx
=11+2xx=0x=1=23
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Alannah Campos
Answered 2022-04-07 Author has 10 answers
You can also use Euler-Maclaurin summation.
The first-order Euler-Maclaurin formula says
i=1nf(i)=1nf(x),dx+{f(1)+f(n)2+1nf(x)(xx12),dx.
Since |xx12|1, with f(x)=2n(n+2x)2 we have
i=1n2n(n+2i)2=1n2n,dx(n+2x)2+Rn,
where |Rn||3f(n)f(1)2|=|3nn(n+2)2|
Therefore,
limni=1n2n(n+2i)2=limn1n2n,dx(n+2x)2
=limn[nn+2x]1n
=limn(13+nn+2)
=23
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