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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 $\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$