Power Sums Limits: Evaluating $\lim_{n\to \infty}$

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In summary, power sums limits are mathematical expressions used to evaluate the limit of a sequence as the number of terms approaches infinity. To evaluate them, the general form of the sequence must be determined and algebraic techniques, such as factoring and using known limit rules, can be used to simplify the expression. The significance of evaluating power sums limits is to understand the behavior of a sequence and find the limit of a series. Some common techniques used to evaluate them include factoring, using known limit rules, and L'Hopital's rule. However, there are restrictions on when power sums limits can be evaluated, such as having a general form that can be simplified and the limit must exist. Some limits may also require more advanced techniques to be evaluated
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Euge
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If ##n## and ##k## are positive integers, let ##S_k(n)## be the sum of ##k##-th powers of the first ##n## natural numbers, i.e., $$S_k(n) = 1^k + 2^k + \cdots + n^k$$ Evaluate the limits $$\lim_{n\to \infty} \frac{S_k(n)}{n^k}$$ and $$\lim_{n\to\infty} \left(\frac{S_k(n)}{n^k} - \frac{n}{k+1}\right)$$
 
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[tex]\frac{S_1(n)}{n}=\frac{n(n+1)}{2n} \rightarrow +\infty [/tex]
[tex]\frac{S_2(n)}{n^2}=\frac{n(n+1)(2n+1)}{6n^2} \rightarrow +\infty [/tex]
[tex]\frac{S_3(n)}{n^3}=\frac{n^2(n+1)^2}{4n^3} \rightarrow +\infty [/tex]
[tex]\frac{S_4(n)}{n^4}=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^4} \rightarrow +\infty [/tex]
[tex]\frac{S_5(n)}{n^5}=\frac{n^2(n+1)^2(2n^2+2n-1)}{12n^5} \rightarrow +\infty [/tex]
In general
[tex]\frac{S_k(n)}{n^k}=1+\frac{1}{k+1}\sum_{j=0}^k \ _{k+1}C_j B_j n^{1-j} =\frac{n}{k+1}+\frac{1}{2}+ o(1/n) \rightarrow +\infty [/tex]
where ##B_j## are Bernoulli numbers.
[tex]\frac{S_k(n)}{n^k}-\frac{n}{k+1} \rightarrow \frac{1}{2} [/tex]
 
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  • #3
Easier is to recognise [itex]1^k + \dots + n^k[/itex] as the upper Darboux sum for [itex]\int_0^n x^k\,dx[/itex] with respect to the partition [itex]\{0, 1, \dots, n\}[/itex]. It follows that [tex]
1^k + \dots + n^k \geq \int_0^n x^k\,dx = \frac{n^{k+1}}{k+1}.[/tex] Hence [tex]
\frac{S_k(n)}{n^k} \geq \frac{n}{k+1} \to \infty.[/tex]
 
  • #4
Sorry, there was a term missing in the limit in the OP. I've made an edit to include the original limit.
 

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