pedrosuavo
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 Feb24 comment How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$? Your solution bugs me because it reminds me of this: $$\sum_{n=1}^{\infty}(-1)^n = -1+1-1+1-1+\cdots = (-1+1)+(-1+1)+(-1+1)+\cdots = 0+0+0+\cdots = 0.$$ Feb22 comment There aren't non-holomorphic polynomials, right? Yeah, that's what I thought. It reminds me of my first Spanish class in which I learned that I didn't need to say "Yo tengo..." since "Tengo..." was sufficient, and that the "Yo" maybe was just overemphasizing the statement. Feb22 asked There aren't non-holomorphic polynomials, right? Jan3 awarded Yearling Dec19 awarded Constituent Dec9 awarded Caucus Nov30 answered Derivatives 1, 2 and 3 and limits Nov20 accepted When Jensen's inequality is equality Nov20 answered When Jensen's inequality is equality Nov20 accepted Sequential compactness in $\mathbb{R}$ Nov20 accepted $f(r) \leq \int_r^{r+1} f(t)dt$ Nov20 asked Fun Lagrange multiplier problem? Oct5 accepted Calculators using Taylor polynomials? Sep25 awarded Tumbleweed Sep24 awarded Autobiographer Sep18 asked Calculators using Taylor polynomials? Aug28 awarded Popular Question Aug6 comment Sequential compactness in $\mathbb{R}$ Maybe I should have used the notation $f^{-1}\{y_n\}$ to make it more clear that it is a pre-image, rather than an inverse. Aug6 comment Sequential compactness in $\mathbb{R}$ @Hayden - Yeah, that actually just occurred to me as I was re-reading my post. Thanks! Aug6 asked Sequential compactness in $\mathbb{R}$