David Spencer
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 Jun 10 awarded Nice Answer Sep 24 awarded Autobiographer Jul 4 awarded Yearling Jul 4 awarded Yearling Jul 19 comment Why do we say the harmonic series is divergent? I certainly intended no offense, and I regret that my wording offended you. As @did says, I was simply checking your understanding, as your second sentence seems to imply (to me) that any constantly increasing series is divergent regardless of whether it tends to infinity at the limit. We can answer most usefully and clearly if we know whether to focus on the terminology or the proof. I hope there are no hard feelings. Jul 18 answered Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer? Jul 18 awarded Commentator Jul 18 awarded Critic Jul 18 comment Determine if it is possible to fit 2 circles in a rectangle Unless I misunderstand, shouldn't it be $d[(r_1,r_1), (l-r_2,h-r_2)] \geq r_1+r_2$? Jul 18 revised Swatting flies with a sledgehammer Fixed typo in title Jul 18 suggested approved edit on Swatting flies with a sledgehammer Jul 18 comment Why do we say the harmonic series is divergent? Do you know what the word "divergent" means here? Jul 18 awarded Editor Jul 18 revised How many rolls until probability of a 5 is at least 1/2? improved clarity Jul 18 comment How many rolls until probability of a 5 is at least 1/2? That would be the probability of rolling a $5$ every time. An event of probability $P$ occurring $n$ times in a row has probability $P^n$, and the probability of the complement (opposite) of $P$ occurring during $n$ trials is $1-P^n$. Jul 18 answered How many rolls until probability of a 5 is at least 1/2? Jul 16 comment Solve for $a$: $V=2(ab+bc+ca)$ Not quite; perhaps I've not explained my hint clearly. Try it this way: I see you refer to "take out the a". Do this from your (correct) expression $\left(\frac{V}{2}\right)-bc=ab+ca$. Jul 16 comment Solve for $a$: $V=2(ab+bc+ca)$ Since you did that one step forwards and then backwards and got something different, you know that you did something wrong there. If you distribute your $2a$ in what you just wrote, you get $\left(\frac{V}{2}\right)-bc=2ab+2ca$. This is very close to what it should be, but the right is a bit different - there's an extra factor of $2$ on each term. That should tell you what you did wrong. Jul 16 comment Solve for $a$: $V=2(ab+bc+ca)$ By backwards, I mean take your expression $\dfrac{\left(\frac{V}{2}\right)-bc}{b+c}=2a$ and multiply it by $b+c$. Do you get the expression you had before you divided by $b+c$? Jul 16 answered Solve for $a$: $V=2(ab+bc+ca)$