« first day (981 days earlier)   

12:27 AM
@MartinSleziak I would say that the 1st and 3rd are near duplicates as well as the 2nd and 4th.
@MartinSleziak Now I need to keep my eyes peeled for others of this same ilk.
@MartinSleziak I would say that we mark the 1st one as a duplicate as the 3rd. Since the 4th one is community wiki, it should probably also stay, with the 2nd one being marked duplicate.
6 hours later…
6:42 AM
@Vladhagen I would base my choice which one to leave open rather on the quality and number of answers, not based on that whether the post is CW. I think that all 4 of them can be considered duplicates, so I should choose one to be left open. I cannot decide between the 3rd one and the 4th one.
In any case I will case my close votes on the first two questions.
Ok, perhaps the 3rd is not a duplicate, since it asks specifically about the approach OP tired....
BTW thanks for your response, Vladhagen!
2 hours later…
8:51 AM
Some more candidates for duplicate questions:
Q: Limit of the sequence $\frac{1^k+2^k+...+n^k}{n^{k+1}}$

Andrei KhHow 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?

Q: Evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}}$

ᴊ ᴀ s ᴏ ɴBy 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 ...

That one was mentioned in robjohn's comment:
This seems to be a duplicate of this question. — robjohn ♦ 11 hours ago
Q: What is the result of $ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $ and why?

DavidWhat 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)?

Q: Proving $\lim_{n \rightarrow \infty} \frac{\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$

PkwssisHow do we prove that $$\lim_{n \rightarrow \infty} \dfrac{\displaystyle\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$$? This type of terms appear in problems on limits, but I am unable to prove this. Please help me out.

I suspect that there might be more copies of this questoin.
3 hours later…
11:42 AM
The only doubt about the four questions above - what does on one of them mean? Should that question be treated differently from the other ones?
We probably have several copies of this question lying around:
Q: $\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluating this limit

Chris's sisEvaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$

Q: Using one limit to compute other

KlobbbyyyI'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?

To one of these two questions was linked this one, which is another duplicate of the four questions I have posted above:
Q: Limit of a Riemann sum $\lim_{n\to \infty} \frac{1^p+2^p+\ldots+n^p}{n^{p+1}}$

vilburI 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...

2 hours later…
1:22 PM
It seems that I was too hasty in voting to close this one:
Sep 14 at 12:28, by Martin Sleziak
Q: Product of all divisors

user67427 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...

Here is link to the review queue: math.stackexchange.com/review/close/264944
BTW I did not know that mods votes are binding in the review queues as well.

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