Kirthi Raman
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 May 7 comment convergence to infinity of$(2^n + 3^n)^{\frac{1}{n}}$ (as $n \to \infty$) @Ignace, Didier pretty much gave you a "big hint" and you don't need anything more to solve this. May 7 comment convergence to infinity of$(2^n + 3^n)^{\frac{1}{n}}$ (as $n \to \infty$) @Ignace I tried to edit as much as I could. (There was no latex at all in your post). May 7 revised convergence to infinity of$(2^n + 3^n)^{\frac{1}{n}}$ (as $n \to \infty$) added 50 characters in body; edited title May 7 comment Solving $xy = a + b\cdot\operatorname{lcm}(x,y) + c\cdot\gcd(x,y)$ given $a,b,c$ Nice! I like it. May 7 comment How many zeroes are in 100! According to WolframAlpha it would be $29$ zeros in $100!$ (trailing $24$ and $5$ zeroes inside), but if you are looking for a method, as Robert Israel said, there is no known method. May 6 comment How to find the removeable or infinite discontinuity of any rational function without the factoring of the polynomial? @Victor you accepted Bill's answer, but yet havn't upvoted. May 6 comment Find $s$ and $t$ with substitution? Follow Andre's answer you get $s=9, t=\frac{25}{4}$ May 6 comment Solving an integral @Jon You have only partially corrected your mistakes. It should be $\sqrt{24}y-1$ and not $\sqrt{24}(y-1)$ May 6 comment $n^2 + 3n +5$ is not divisible by $121$ You got a great answer by Bill, when a number is divisible by $121$ what is it congruent to mod $11^2$? May 6 comment The maximum number of nodes in a binary tree of depth $k$ is $2^{k}-1$, $k \geq1$. @rajansthapit what is the name of tile/author of the book? Gadi has the correct answer. May 6 answered Derivative of $e^{-x}$ May 6 comment The Importance Of Good Teachers and Guidance In the Academics Today at least there are tons of resources as an alternative, assuming you are self-motivated. If a teacher is not good, it really brings down the confidence of a student (Personally from what I noticed in my Son's case). May 6 comment If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. My +1 and Thanks. May 6 comment If $b^2$ is the largest square divisor of $n$ and $a^2 \mid n$, then $a \mid b$. @BillDubuque Recommend the long line to be split as \begin{align*} a^2 | b^2c \Rightarrow a^2 | (bc)^2 &\Rightarrow a^2 | a^2c, abc, b^2c \\ &\Rightarrow a^2 | (a,b)^2c \Rightarrow (a/(a,b))^2 | c \end{align*} My +1 for this answer. May 6 comment Result of the product $0.9 \times 0.99 \times 0.999 \times …$ @PaulManta this converges to $0.89$, but you have to think how to show that. May 6 comment Result of the product $0.9 \times 0.99 \times 0.999 \times …$ @PaulManta I changed the \cdots to \times because it was confusing with the numbers themselves with decimal values. May 6 revised Result of the product $0.9 \times 0.99 \times 0.999 \times …$ edited title May 6 comment Result of the product $0.9 \times 0.99 \times 0.999 \times …$ How about $(1-\frac{1}{10})(1-\frac{1}{100})...= \prod_{i=0}^{\infty}(1-\frac{1}{10^i})$ May 6 awarded Necromancer May 5 awarded Nice Answer