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Evaluate the limit with solution (8 pts.):
[tex]\lim_{n \to \infty} \frac{1}{n^4}[(\sum_{k=1}^{n}k) + 2(\sum_{k=1}^{n-1}k) + 3(\sum_{k=1}^{n-2}k) + ... + n][/tex]

Sagot :

Nzmidn

SOLUTION:

[tex]\sf \lim_{n \to \infty} \frac{1}{n^4}\bigg[\Big(\sum\limits_{k=1}^{n}k\Big)+ 2\Big(\sum\limits_{k=1}^{n-1} k\Big) + 3\Big(\sum\limits_{k=1}^{n-2}k\Big) + ... + n\bigg][/tex]

[tex]\sf \lim_{n \to \infty} \frac{1}{n^4}\bigg[\sum\limits_{x=1}^{n}\sum\limits_{k=1}^{n-x+1} kx \bigg][/tex]

[tex]\sf \lim_{n \to \infty} \big(\frac{1}{n^4}\big)\sum\limits_{x=1}^{n}x\big[\frac{1}{2} (n-x+1)(n-x+2)\big][/tex]

[tex]\sf \lim_{n \to \infty} \frac{1}{2} \big(\frac{1}{n^4}\big)\sum\limits_{x=1}^{n} (n^2 + 3n + 2)x - (2n+3)x^2 + x^3[/tex]

[tex]\sf \lim_{n \to \infty} \frac{1}{2} \big(\frac{1}{n^4}\big)\big[\frac{1}{2}n(n+1)(n^2+3n+2)-\frac{1}{6}n(n+1)(2n+1)(2n+3) + \frac{1}{4}n^2(n+1)^2\big][/tex]

[tex]\sf \frac{1}{2}\big(\frac{1}{n^4})(\frac{1}{2}n^4-\frac{1}{6}n^4(4) + \frac{1}{4}n^4)[/tex]

[tex]\sf \frac{1}{2}\big(\frac{1}{\cancel{n^4}}\big)(\frac{1}{2}\cancel{n^4}-\frac{2}{3}\cancel{n^4}+\frac{1}{4}\cancel{n^4})[/tex]

[tex]\sf \frac{1}{2}(\frac{1}{12})[/tex]

[tex]\sf \boxed{\sf \frac{1}{24} }[/tex]

ANSWER:

1/24

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