Addendum No. 2 1: Fill & Download for Free

HO boy. I have got some really nice tricks that both have major risk involved, but trust me. The payoff is IMMENSE.First order of business: Fall damage reducersThere are some pretty neato ones that I got.1. The Slime BlockBoing!Its description states that “Landing on a slime block does not cause fall damage unless the player is sneaking.” And also “A player or mob that falls onto the top of a slime block bounces to a height proportional to the falling velocity. The bounce height quickly deteriorates.”. These two things combined means that if you land on one you bounce without taking any damage, while also nerfing any damage taken when you hit the ground after the first bounce. So, say you jump off of a 225 block high cliff; you would only take damage when you fall from the 50 blocks that it would bounce you up to. Another thing, slime blocks are instantly breakable with any tool. So that means that if you’re being chased and your chaser thinks they can get away with jumping on your slime block after you, you can just yoink it out from under their feet.2. The Hay BlockHay is for horses…in a more literal senseBasically, the hay block is a literal hay block. You take straw and combine them to mak hay. Not much more to say than that.Unless, of course, if Microsoft had programmed a fun hidden feature into it.You see, hay blocks give the player a big cushion when they fall on it. This means that the 225 block fall will kill you, but the damage you’ll take will technically have been not as bad.3. And last, but probably not least, the bedSoft and cushiony…just the way I like it.A bed does the same thing a slime does, but does a worse job of doing it. What a surprise. When you jump on it too hard, you die. Deep lore. In all fairness, though, it does its job well enough.Addendum No. 2; Off-shore tricksLily PadsInsert funny quit here…cause I can’tIt’s a lily pad. Place it on water and you can walk there. Boom. Like Jesus (or something like that).2. DolphinsAw, how cute!If you get lucky while diving, you’ll get a boost from one of these. Find multiple and the boost stacks.Addendum No. 2; Part 2; Lava is really hot.No, like, really really hot.Boat MLGBasically lily pads, but you waste a lot of wood. It’s funny to trick your comrades into thinking you’re gonna die and then whipping out the boats.2. StridersThey walk on lava. RIP striders. You’ll be missed.Sorry for the degrade in quality. Anyways, I’ll sign off with this.“Now, we are all sons of bitches” (Kenneth Bainbridge)Ian Leath

We first compute this sum, assuming convergence (see addendum below for its details).Noting that [math]e^{2it} = \cos(2t) + i \sin(2t)[/math] by Euler’s Identity, we consider the more general sum[math]\displaystyle S = \sum_{n=1}^{\infty} \frac{e^{2in}}{n}. \tag*{}[/math]By using basic manipulations, we obtain[math]\begin{align*} S &= \displaystyle \sum_{n=1}^{\infty} \int_0^x t^{n-1} \, dt \Bigg|_{x = e^{2i}}\\ &= \displaystyle \int_0^x \frac{1}{1-t} \, dt \Bigg|_{x = e^{2i}}, \text{ via geometric series}\\ &= \displaystyle -\text{Log}(1 - e^{2i})\end{align*} \tag*{}[/math]Next, we use the definition of the (complex) logarithm:[math]\begin{align*} S &= \displaystyle -\text{Log}(1 - \cos{2} - i \sin{2})\\ &= \displaystyle -\Big(\ln|1 - \cos{2} - i \sin{2}| + i \, \text{Arg}(1 - \cos{2} - i \sin{2}) \Big)\\ &= \displaystyle -\Big(\ln\sqrt{(1 - \cos{2})^2 + \sin^2{2}} + i \arctan\Big(\frac{-\sin{2}}{1 - \cos{2}}\Big)\Big)\\ &= \displaystyle -\frac{1}{2}\ln(2 - 2\cos{2}) + i \arctan\Big(\frac{\sin{2}}{1 - \cos{2}}\Big). \end{align*} \tag*{}[/math]Taking imaginary parts of both sides yields[math]\displaystyle \sum_{n=1}^{\infty} \frac{\sin(2n)}{n} = \arctan\Big(\frac{\sin{2}}{1 - \cos{2}}\Big) = \arctan(\cot{1}) = \frac{\pi}{2} - [/math][math]1[/math][math]. \tag*{}[/math]Remark: Taking real parts of both sides yields the companion result[math]\displaystyle \sum_{n=1}^{\infty} \frac{\cos(2n)}{n} = -\frac{1}{2}\ln(2 - 2\cos{2}) = -\frac{1}{2}\ln(4 \sin^2{1}) = -\ln(2 \sin{1}). \tag*{}[/math](Double checked the decimal approximations on Wolfram Alpha.)Addendum:By popular demand, we now show the convergence of[math]\displaystyle S = \sum_{n=1}^{\infty} \frac{e^{2in}}{n} \tag*{}[/math]by using Dirichlet’s Test.Let [math]a_n = \frac{1}{n}[/math] and [math]b_n = e^{2in}[/math].First of all, it is clear that [math]\{a_n\}[/math] is a decreasing sequence of real numbers that converges to 0.Next, we show that [math]|\sum_{n=1}^N b_n|[/math] is bounded for all positive integers [math]N[/math]. By using the finite geometric series,[math]\begin{align*} \displaystyle \Big|\sum_{n=1}^N b_n \Big| &= \displaystyle \Big|\frac{e^{2i} (e^{2iN} - 1)}{e^{2i} - [/math][math]1[/math][math]} \Big| \\ &\leq \displaystyle \frac{|e^{2i}| (|e^{2iN}| + |1|)}{|e^{2i} - [/math][math]1[/math][math]|} \\ &= \displaystyle \frac{1 ([/math][math]1[/math][math] + 1)}{|e^{i} (e^{i} - e^{-i})|} \\ &= \displaystyle \frac{2}{1 \cdot |2i \sin{1}|} \\ &= \csc{1}.\end{align*} \tag*{}[/math]Therefore, [math]S[/math] is convergent by Dirichlet’s Test.

Many people have already posted how differentiating both sides of the equation[math]\displaystyle\dfrac{1}{1-x}=\sum_{n=0}^\infty x^n\tag{$\dagger$}[/math]twice gives you the answer you need. Indeed, I’d say that these people’s approach is the best one. Here, I’ll describe a different approach which does not require any derivatives. We assume, before we start, that our infinite series is convergent.Let[math]S=x+4x^2+9x^3+16x^4+\cdots\tag{1}[/math]which is, of course, our sum.Multiplying [math]([/math][math]1[/math][math])[/math] by [math]x[/math], we have:[math]xS=x^2+4x^3+9x^4+16x^5+\cdots.\tag{2}[/math]Thus, by finding [math](1)-(2)[/math],[math](1-x)S=x+3x^2+5x^3+7x^4+9x^5+\cdots.\tag{3}[/math]Now multiply [math](3)[/math] by [math]x[/math]:[math]x(1-x)S=x^2+3x^3+5x^4+7x^5+\cdots.\tag{4}[/math]Finding [math](3)-(4)[/math], we obtain:[math]\big((1-x)-x(1-x)\big)S=x+2x^2+2x^3+2x^4+2x^5+\cdots[/math][math](1-x)(1-x)S=x+2x^2(1+x+x^2+x^3+\cdots)[/math][math](1-x)^2S-x=\dfrac{2x^2}{1-x}[/math], by [math](\dagger)[/math] (or by summing up the geometric series)[math](1-x)^2S=x+\dfrac{2x^2}{1-x}[/math][math](1-x)^2S=\dfrac{x(1-x)+2x^2}{1-x}[/math][math](1-x)^2S=\dfrac{x(1+x)}{1-x}[/math][math]S=\dfrac{x(1+x)}{(1-x)^3}.\tag{$\star$}[/math]Equation [math](\star)[/math] is the expression we wanted to find.Addendum: Technically, the above technique may be used in the context of formal power series, where the end result may be written as[math]\displaystyle(1-x)^3\left(\sum_{n=0}^\infty n^2x^n\right)=x(1+x).[/math]The above equation is true in the ring of formal power series (with integer coefficients), whether or not the series actually converges.