How would someone find the limit of the sequence $a_n = \frac{1^k+2^k+...+n^k}{n^{k+1}}, k \in \mathbb{N}$ as $n$ goes to Infinity? Can someone give me maybe a hint where to start?
By considering: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^1}{n^{2}} = \frac 1 2$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^{3}} = \frac 1 3$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^3}{n^{4}} = \frac 1 4$$ Determine if this is true: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac ...
What is the result of the next limit: $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$ Why (theorem)?
Evaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$
I've calculated $\lim_{n\to\infty}\dfrac{1^p+2^p+\cdots+n^p}{n^{p+1}}=\dfrac1{p+1}$ where $p\in\mathbb{N}$ fixed. I feel it should help me get this one $\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\cdots+n^p}{n^{p}}-\dfrac{n}{p+1}\right)$, but I'm not sure how. Any hints?
I have$$\lim_{n\to \infty} \frac{1^p+2^p+\ldots+n^p}{n^{p+1}}=$$ I managed to simplify it down to $$=\lim_{n\to \infty}\left( \left(\frac1n\right)^p \cdot \frac1n + \left(\frac2n\right)^p \cdot \frac1n + \ldots + \left(\frac{n}{n} \right)^p \cdot \frac1n \right)=$$ $$=\lim_{n\to \infty} \sum_{i=1...
Prove that $\prod_{d \mid n}d=n^{v(n)/2}$ where $v(n)$ is the sum of divisors function. We have if $n=p_{1}^{a_{1}}p_{2}^{a_{2}} \dots p_{k}^{a_{k}}$ then $v(n)=(a_{1} +1)(a_{2}+1) \dots (a_{k} +1)$ substituting this in the expression does not reach anything, is there any way to express $\p...
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