MHB Computing the Limit of a Power Series

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The limit of the power series is computed as follows: the expression $\lim_{n\to +\infty}\dfrac{1^p+2^p+\cdots +n^p}{n^{p+1}}$ can be evaluated using the Stolz-Cesaro theorem, leading to the result $\frac{1}{p+1}$. Additionally, this limit can be interpreted as the limit of Riemann sums for the function $f(x) = x^p$ over the interval [0,1], which corresponds to the integral $\int_0^1 x^p \, dx$. The integral evaluates to the same result, confirming that the limit is indeed $\frac{1}{p+1}$. The discussion highlights both the Stolz-Cesaro theorem and the Riemann sum approach as valid methods for calculating the limit.
Julio1
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Compute $\displaystyle\lim_{n\to +\infty}\dfrac{1^p+2^p+3^p+\cdots +n^p}{n^{p+1}}.$
Hello!, how it can calculate this limit? Thanks :)
 
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I am not sure what techniques you are familiar with for calculating limits, but this seems to be a job for Stolz-Cesaro theorem :

$$\begin{aligned}\lim_{n \to \infty} \frac{1^p + 2^p + \cdots + n^p}{n^{p+1}} &= \lim_{n \to \infty} \frac{\left ( 1^p + 2^p + \cdots + (n+1)^p \right ) - \left( 1^p + 2^p + \cdots + n^p \right)}{(n+1)^{p+1} - n^{p+1}} \\ &= \lim_{n \to \infty} \frac{(n+1)^p}{(n+1)^{p+1} - n^{p+1}} \\ & = \lim_{n \to \infty} \frac{(n+1)^p}{(p+1) \cdot n^p + \mathcal{O}(n^{p-1})} \\ &= \lim_{n \to \infty} \frac{1}{p+1} \cdot \frac{1}{1+o(1)} = \boxed{\dfrac1{p+1}}\end{aligned}$$

Where $\mathcal{O}$ and $o$ are asymptotic notations. I am guessing that there might be some elementary way to do to this, so am interested in other solutions.
 
Last edited:
Julio said:
Compute $\displaystyle\lim_{n\to +\infty}\dfrac{1^p+2^p+3^p+\cdots +n^p}{n^{p+1}}.$
Hello!, how it can calculate this limit? Thanks :)

I'm assuming $p$ is a real number different from $-1$. The limit, which can be written

$$\lim_{n\to +\infty} \frac{1}{n}\sum_{k = 1}^n \left(\frac{k}{n}\right)^p,$$

is the limit of a sequence of Riemann sums for the function $f(x) = x^p$ over the interval $[0,1]$. So it has value

$$\int_0^1 x^p \, dx.$$

Compute the integral to get the result.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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