query: Why does an argument similiar to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$ show that $2+4+8+...=-2$
positive_passages.docid: 13363
positive_passages.text: Why does an argument similiar to 0.999...=1 show 999...=-1?
negative_passages.docid: 27969
negative_passages.text: Why does $\frac{1}{x} < 4$ have two answers?
negative_passages.docid: 143236
negative_passages.text: $ S_{n}=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+...........+\frac{x^{2^{n}}}{(x+1)(x^2+1)...(x^{2^{n}}+1)}$
negative_passages.docid: 61782
negative_passages.text: How to prove that $f(x) = \frac{-2x+1}{(2x-1)^2-1}$ is one-to-one on $(0,1)$?
negative_passages.docid: 93756
negative_passages.text: $1+\frac{1}{2} +\frac{1}{3} +...+\frac{1}{p-1} =\frac{a}{b}$
negative_passages.docid: 19561
negative_passages.text: Why is $\frac{1}{\frac{1}{X}}=X$?
negative_passages.docid: 62072
negative_passages.text: Is $2+2=4$ an identity?
negative_passages.docid: 63553
negative_passages.text: Show that $\frac{1}{(n+1)!}(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots)=\frac{1}{n!n}$
negative_passages.docid: 102334
negative_passages.text: How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
negative_passages.docid: 56059
negative_passages.text: If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
negative_passages.docid: 36437
negative_passages.text: How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$
negative_passages.docid: 26474
negative_passages.text: how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$?
negative_passages.docid: 4340
negative_passages.text: Why does $1+2+3+\cdots = -\frac{1}{12}$?
negative_passages.docid: 125274
negative_passages.text: For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$
negative_passages.docid: 109365
negative_passages.text: Prove that 1/2 + 1/4 + 1/8 ....... = 1
negative_passages.docid: 39908
negative_passages.text: Finding the error in this proof that 1=2
negative_passages.docid: 101491
negative_passages.text: If $\frac{a}{|z_2-z_3|}=\frac{b}{|z_3-z_1|}=\frac{c}{|z_1-z_2|}$, then why is $\frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2}=0$?