Oi, alguém pode me ajudar com essa questão?
Agradeço desde já.
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Olá, Eletronpaulo.
![s(n)=\sum\limits_{i=1}^n\frac{3i+3}{n^2}=\frac3{n^2}\sum\limits_{i=1}^n( i+1)=\frac3{n^2}(\sum\limits_{i=1}^n i+\sum\limits_{i=1}^n 1)=\\\\=\frac3{n^2}[\underbrace{(1+2+...+n)}_{\text{PA de raz\~ao 1}}+\underbrace{(1+...+1)}_{n\text{ vezes}}]=\frac3{n^2}[\frac{n(n+1)}2+n]=\\\\=\frac3{n^2}[\frac{n^2+n}2+n]=\frac3{n^2}[\frac{n^2}2+\frac n2+n]=\frac32+\frac3{2n}+\frac3 n\\\\\therefore\boxed{\lim\limits_{n\to\infty}s(n)=\frac32}](https://tex.z-dn.net/?f=s%28n%29%3D%5Csum%5Climits_%7Bi%3D1%7D%5En%5Cfrac%7B3i%2B3%7D%7Bn%5E2%7D%3D%5Cfrac3%7Bn%5E2%7D%5Csum%5Climits_%7Bi%3D1%7D%5En%28+i%2B1%29%3D%5Cfrac3%7Bn%5E2%7D%28%5Csum%5Climits_%7Bi%3D1%7D%5En+i%2B%5Csum%5Climits_%7Bi%3D1%7D%5En+1%29%3D%5C%5C%5C%5C%3D%5Cfrac3%7Bn%5E2%7D%5B%5Cunderbrace%7B%281%2B2%2B...%2Bn%29%7D_%7B%5Ctext%7BPA+de+raz%5C%7Eao+1%7D%7D%2B%5Cunderbrace%7B%281%2B...%2B1%29%7D_%7Bn%5Ctext%7B+vezes%7D%7D%5D%3D%5Cfrac3%7Bn%5E2%7D%5B%5Cfrac%7Bn%28n%2B1%29%7D2%2Bn%5D%3D%5C%5C%5C%5C%3D%5Cfrac3%7Bn%5E2%7D%5B%5Cfrac%7Bn%5E2%2Bn%7D2%2Bn%5D%3D%5Cfrac3%7Bn%5E2%7D%5B%5Cfrac%7Bn%5E2%7D2%2B%5Cfrac+n2%2Bn%5D%3D%5Cfrac32%2B%5Cfrac3%7B2n%7D%2B%5Cfrac3+n%5C%5C%5C%5C%5Ctherefore%5Cboxed%7B%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7Ds%28n%29%3D%5Cfrac32%7D "s(n)=\sum\limits_{i=1}^n\frac{3i+3}{n^2}=\frac3{n^2}\sum\limits_{i=1}^n( i+1)=\frac3{n^2}(\sum\limits_{i=1}^n i+\sum\limits_{i=1}^n 1)=\\\\=\frac3{n^2}[\underbrace{(1+2+...+n)}_{\text{PA de raz\~ao 1}}+\underbrace{(1+...+1)}_{n\text{ vezes}}]=\frac3{n^2}[\frac{n(n+1)}2+n]=\\\\=\frac3{n^2}[\frac{n^2+n}2+n]=\frac3{n^2}[\frac{n^2}2+\frac n2+n]=\frac32+\frac3{2n}+\frac3 n\\\\\therefore\boxed{\lim\limits_{n\to\infty}s(n)=\frac32}")
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