See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$  $x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$  $2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$  Then:  $x=1$  Now I use the same argument to prove $2+4+8+...=-2$  $x=2+4+8+...$  $2x=4+8+16+...$  Then:  $x=-2$
46383
I accept that two numbers can have the same supremum depending on how you generate a decimal representation. So $2.4999\ldots = 2.5$ etc. Can anyone point me to resources that would explain what the below argument that shows $999\ldots = -1$ is about? Here is the most usual proof I see that $0.999\ldots = 1$:  $x=0.999\ldots$ $10x=9.999\ldots$ $10x - x = 9$ $x=1$  Using this same argument template I can show $999\ldots=-1$:  $x= \ldots9999.0 $ $0.1x= \ldots9999.9$ $0.1x - x = 0.9$ $x=-1$  What might this mean? Edit from one of the comments: $$\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1$$
115203
Prove that for $n \in \mathbb{N}$  (a) $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\geq \frac{2}{3}$  And   (b) $\frac{1}{2} \leq \frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}+....&lt; \frac{2}{3}$  my attempt :  $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\\ &gt;\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}\\ =\frac{n+1}{2n}&gt;\frac{1}{2}$  Where i am missing and how to processed second question
35641
I have the following problem:  Prove that if $k &gt; 1$, then $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2}$$ Hence simplify $$\sum\limits_{k = 2}^n \dfrac{k}{(k^2 - 1)^2}$$  My Work: $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k - 1)^2 (k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k^2 - 2k + 1)(k^2 + 2k + 1)}$$ I could have continued and multiplied the denominator out, but, at this point, I'm thinking that I'm actually expected to have proceeded differently, rather than multiplying the denominator out fully? And I don't have solutions for this problem, so I can't check anything. I would appreciate it if someone could please take the time to explain this problem.
72148
Proving that the series 1 + ... + $1 / \sqrt{x}$ &lt; $2 \sqrt{x}$  I am doing it by simple induction adding $1/\sqrt{x+1}$ to both sides, but I can't find a way to manipulate this expression and find that the new series is $&lt; 2 \sqrt{x+1}$.  Can someone show me the correct process?  I failed many times and the furthest I've gotten has been to prove that it's $&lt; 2 \sqrt{x+1} + 1$ which is close, but not enough.
5951
I am currently studying proving by induction but I am faced with a problem.  I need to solve by induction the following question.  $$1+2+3+\ldots+n=\frac{1}{2}n(n+1)$$      for all $n &gt; 1$.  Any help on how to solve this would be appreciated.    This is what I have done so far.  Show truth for $N = 1$   Left Hand Side = 1  Right Hand Side = $\frac{1}{2} (1) (1+1) = 1$  Suppose truth for $N = k$  $$1 + 2 + 3 + ... + k = \frac{1}{2} k(k+1)$$  Proof that the equation is true for $N = k + 1$  $$1 + 2 + 3 + ... + k + (k + 1)$$  Which is Equal To  $$\frac{1}{2} k (k + 1) + (k + 1)$$  This is where I'm stuck, I don't know what else to do. The answer should be:  $$\frac{1}{2} (k+1) (k+1+1)$$  Which is equal to:  $$\frac{1}{2} (k+1) (k+2)$$  Right?  By the way sorry about the formatting, I'm still new.
118928
Prove that $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$$   I tried to rationalize the denominator but I always end up with a large fraction that won't cancel out. Is there something I'm missing?  Thanks in advance
86636
If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.  I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that $ \frac{-1}{2}\le\ ab+bc+ca$ .  Thanks in advance !
13515
Evaluate $$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(4x)} + \dfrac{1}{\sin(8x)} + \dfrac{1}{\sin(16x)}$$  It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed. Rewriting and using trigonometric identities $$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(2x) \cos (2x)} + \dfrac{1}{ 2\big [2\sin (2x)\cos (2x)\cos (4x)\big ]} + \dfrac{1}{\sin(16x)}$$ What am I missing? Regards
45865
Are there any positive integers $x,y,z$ such that $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1?$$ Prove/Disprove.   I've plugged in random positive integers for $x,y,z$ and I have not been able to get the equation to equal $1$.
116754
I had the equation: $$\frac{1}{(1+x)(1-x)^2}=\frac{A}{(1-x)^2}+\frac{B}{1-x}+\frac{C}{1+x}$$ $$A=\left.\frac{1}{1+x}\right|_{x=1} = \frac{1}{2} \tag2$$ What does $(2)$ mean? Particularly, the notation $|_{x=1}$
5998
Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} &gt; \frac{13}{24}$ for $n&gt;1,n\in\Bbb N$  To solve it I used induction but it is leading me nowhere my attempt was as follows:  Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$  so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} &gt; 13/24$  I don't know where to go from here please help.
29195
I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?
76163
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.   I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see how that helps.
1937
How to prove for $x&gt;0$, $\dfrac{x}{1+x^2}&lt;\arctan x&lt;x$?   I started with saying $0&lt;\arctan x&lt;\frac{\pi}{2}$  when $x&gt;0$  I'm just not sure how to proceed with this proof, it works when I take random  values but I cant see how to prove that it works for any $x$.
109392
Prove by induction $\sum \frac {1}{2^n} &lt; 1$  Well supposing the base case has been shown to be true, I start with the induction step:  Suppose true for n = k:  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} &lt; 1$$  Want to show this is true for:  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} &lt; 1$$  Now this is where i am getting stuck, should i be tryig to show that  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} &lt; 1 +\frac{1}{2^{k+1}}$$ or should i be attempting to show that  $1 + \frac{1}{2^{k+1}} &lt; 1$ which is utter nonsense. So i am stuck on what exactly to prove here.  Note i also thought of maybe using the geometric series of $\frac {1}{2} $ as some upper bound but then that would provide me a value that was greater than 1 so it didn't work out
910
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.
112982
Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{(2x)}  \, dx = \frac{\ln|2x|}{2}+ C$$   Check: By taking derivative we get $\frac{1}{x}$  Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{2x}  \, dx = \frac{1}{2}\int  \frac{1}{x}  \, dx =\frac{\ln|x|}{2}+ C$$  Check: By taking derivative we get $\frac{1}{x}$  Question: Is  $\int  \frac{1}{2x}  \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$?