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 Yearling
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Apr
25
comment Show that the limit $\displaystyle \lim_{n\to \infty }\frac{a_{n}}{n}$ exists.
Are you sure "a sequence of real numbers"? E.g. $a_{n}=-n^2$ and $-(n+m)^2 < -n^2 - m^2 +1$
Apr
25
comment Show that the limit $\displaystyle \lim_{n\to \infty }\frac{a_{n}}{n}$ exists.
A rectification is required $a_n=b_n - d$ ...
Apr
13
comment Can we replace the upper limit condition of the Sieve of Eratosthenes $\sqrt{n}$ with the value $\sqrt{p}$ where $p$ is the last sieved prime $\lt n$?
Yep, but I couldn't omit this case, for completeness ;) Also, my previous comment re Andrica's conjecture would suggest $\sqrt{p_{\pi(q^2)}} < q < \sqrt{p_{\pi(q^2)+1}}$ and $\sqrt{p_{\pi(q^2)+1}} - \sqrt{p_{\pi(q^2)}} < 1$ then $\left \lceil \sqrt{p_{\pi(q^2)}} \right \rceil=q$
Apr
13
comment Can we replace the upper limit condition of the Sieve of Eratosthenes $\sqrt{n}$ with the value $\sqrt{p}$ where $p$ is the last sieved prime $\lt n$?
I think that's not a counter example, but rather an example because you are considering $101$ as the lower bound and not any $p$ lower than $p \leq \sqrt{10193}=100.9 < 100+1$
Apr
12
answered Can we replace the upper limit condition of the Sieve of Eratosthenes $\sqrt{n}$ with the value $\sqrt{p}$ where $p$ is the last sieved prime $\lt n$?
Apr
12
comment Can we replace the upper limit condition of the Sieve of Eratosthenes $\sqrt{n}$ with the value $\sqrt{p}$ where $p$ is the last sieved prime $\lt n$?
There could be more than one prime between $[\frac{n}{2}, n]$, ramanujan.sirinudi.org/Volumes/published/ram24.html
Apr
11
comment Can we replace the upper limit condition of the Sieve of Eratosthenes $\sqrt{n}$ with the value $\sqrt{p}$ where $p$ is the last sieved prime $\lt n$?
Why do you care about using $\sqrt{p}$ rather than $\sqrt{n}$? The closest prime to $n$, is $p_{\pi(n)}$, basically $p_{\pi(n)} \leq n < p_{\pi(n)+1}$. Assuming Andrica's conjecture is true $\sqrt{n} - \sqrt{p_{\pi(n)}} < \sqrt{p_{\pi(n)+1}} - \sqrt{p_{\pi(n)}}<1$. I can't say there is a big performance gain in such a replacement.
Apr
10
revised Representing $ 2 i \sin(2 \pi n z) $ as a product
added 1 character in body
Apr
8
suggested rejected edit on Representing $ 2 i \sin(2 \pi n z) $ as a product
Apr
8
revised Representing $ 2 i \sin(2 \pi n z) $ as a product
added 116 characters in body
Apr
8
answered Representing $ 2 i \sin(2 \pi n z) $ as a product
Apr
7
comment Limit of n*sin(1/n) as n goes to infinity
See some ideas here math.stackexchange.com/questions/75130/…
Apr
6
comment Limit of n*sin(1/n) as n goes to infinity
Have you used mean value theorem?
Apr
3
revised Finding an upperbound for $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$
added 177 characters in body
Apr
3
comment Finding an upperbound for $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$
Yep, $\varepsilon $ is a constant ...
Apr
3
comment Finding an upperbound for $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$
$\prod_{k=1}^{n} \frac{p_{k}-1}{p_{k}} \sim \frac{ e^{-\gamma }}{\ln{n}} \Leftrightarrow \lim_{n \to \infty } \frac{\prod_{k=1}^{n} \frac{p_{k}-1}{p_{k}}}{\frac{ e^{-\gamma }}{\ln{n}}}=1 $ which means $\exists \varepsilon >0$ such that $\prod_{k=1}^{n} \frac{p_{k}-1}{p_{k}} < (1+\varepsilon ) \frac{ e^{-\gamma }}{\ln{n}} $ always!
Apr
3
comment Finding an upperbound for $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$
Just make sure you deal with the case $k=1$, e.g. $\prod_{k=2}^{n} \frac{p_{k}-1}{p_{k}} = 2 \cdot \prod_{k=1}^{n} \frac{p_{k}-1}{p_{k}} \sim \frac{2 \cdot e^{-\gamma }}{\ln{n}}$
Apr
3
comment Finding an upperbound for $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$
The best I can think of then is $\prod_{k=2}^{i} \frac{p_{k}-2}{p_{k}} < \prod_{k=2}^{i} \frac{p_{k}-1}{p_{k}} \sim \frac{e^{-\gamma }}{\ln{n}}$ which is Merten’s theorem (section 22.8 in this book matematica.cubaeduca.cu/medias/pdf/842.pdf).
Apr
3
comment Finding an upperbound for $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$
How about $\sum_{i=2}^{n} \frac {1}{p_i} < f(n)< \frac{n}{3}$ ?
Apr
3
answered Finding an upperbound for $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$