I saw $\zeta (1/2)=-1.4603545088…$ in this link. But how can that be? Isn’t $\zeta (1/2)$ divergent since $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..$ ?

The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4).$ Find 3 more solutions in positige integers. [hint. look for solutions of the form $(a,b,c) = (xz, yz, z^2)$ Attempt: So I tried to use the hint in relation to the triple that they gave that worked. So I observed that […]

In the book The elementary proof of Prime Number Theorem it says that The prime number theorem $\psi (x)\sim x$ is equivalent to $\int_1^\infty \frac{\psi (t)-t}{t^2} =-\gamma -1$, where $\psi (x)=\sum_{n\le x} \Lambda (n)$, $\gamma $ is the Euler constant. The book also gives a hint to prove $\sum_{n\le x}\frac{\Lambda (n)}{n}=\int_1^x \frac{\psi(t)}{t^2} dt+\frac{\psi (x)}{x}$ first, […]

For an integer $n \geq 1$, verify the formula: $\sum\limits_{d|n} \mu (d) \lambda(d)= 2^{\omega (n)}$ I know that this problem somehow uses the Mobius Inversion Formula but I am VERY confused how to use it.

Question Let $p_r$ be the $r’th$ prime. Is it true that, $$\sum_{r=1}^\infty s^r \ln(p_r) \sim \frac{s}{(1-s)} $$ I know this looks bizarre but kindly consider the argument below. I’m also interested in knowing if this technique (of summing to simplify via asymptotics) is known? Argument I recently observed an interesting behavior of the following series: […]

Find all integral solutions of the equation $a+b+c=abc$. Is $\{a,b,c\}=\{1,2,3\}$ the only solution? I’ve tried by taking $a,b,c=1,2,3$.

Given two integers $N$ and $M$ , How to find out number of arrays A of size N, such that : Each of the element in array, $1 ≤ A[i] ≤ M$ For each pair i, j ($1 ≤ i < j ≤ N$) $GCD(A[i], A[j]) = 1$ $M$ can be at max $100$. But […]

Possible Duplicate: On the sequence $x_{n+1} = \sqrt{c+x_n}$ I am wondering how many different solutions one can get to the following question: Calculate $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ Post your favorite solution please.

When we add the digits of number for eg $2478125$ $$2+4+7+8+1+2+5=29;\\ 2+9=11;\\ 1+1=2$$ and when we square the digits and add them following this rule $$2^2+4^2+7^2+8^2+1^2+2^2+5^2=163;\\ 1^2+6^2+3^2=46; \\ 4^2+6^2=52;\\ 5^2+2^2=29;\\ 2^2+9^2=85;$$and this goes on, so the question is will it ever reach a single digit.And what is the proof of it.

Let $\gcd(b,6) = 1$. Prove that $b^2 \equiv 1 $ (mod 24). Now I have that as $\gcd(b,6) = 1$, we know that $3\nmid b $ and $2\nmid b$ (else the GCD would be 3 or 2 resp.) So as $2\nmid b$, $b$ must be odd. Hence $b^2$ is also odd. Then I’m not sure […]

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