One of my friends has given me the following problem:     $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{\infty} = ?$$   He said that the answer is $1$. He has given his argument as that this is a part of the whole, and if we add together all the part, then we'd have a $1$. But said, disagreed with him that it'd not be $1$, rather it'd be $\frac{\infty - 1}{\infty}$. I'm not sure if this is correct. But I have given the argument with this view:   If we add together only the first two fractions, that is $\frac{1}{2}$ and $\frac{1}{4}$, then the sum will be $\frac{3}{4}$, which is equal to $\frac{4-1}{4}$. Likewise, if we add together the first three, four, five, then the sum will be $\frac{x-1}{x}$, where $x$ is the denominator of last fraction. These happen because every time divide ramainder(whole in the first case) in the equal halves, there remain one part of the equal halves and after we add together all the part(follows the pattern, never break the pattern) there remain $\frac{1}{x}$ of the whole part. So we can nonetheless say this that this only became 1(the whole), if we add add the remainder in the sum, that's mean there will be $2$ of the last fraction. But that will break the pattern. But he argued with me and wanted me to ask the question here. I also said that if we round it then it will be $1$, rather it never will be $1$.    So, what is your opinion about it? If included both, argument and mathematical calculation, then that will be great.
121550
There is a couple of limits that I failed to find:  $$\lim_{n\to\infty}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}$$  and  $$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n-1}}{3^{n-1}}$$  There is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result.  So my questions are:   What should I do when I need a limit of infinite sum? (are there any rules of thumb?) What theorems or topics from calculus should I know to solve these problems better?   I am new to math and will appreciate any help. Thank you!
7610
Why multiplying fractions is equal to multiply the tops, multiply the bottoms? $$\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b \times d},$$ And why $$\frac{a}{b}\times \frac{c}{c}=\frac{a}{b},$$ Also why $$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}.$$ I understand it, but I want a mathematical approach as a math student proves it. Also I want to know the mathematics topic of this question (number theory, logic, etc). A full answer is not necessary. Just a reference.
29310
Find the sum to $n$ terms as well as the sum to infinity of the series:  $$\frac{1}{3} + \frac{1}{4\cdot 2!} + \frac{1}{5\cdot 3!} +\cdots.$$  I was trying this question many times but I didn't get the answer. I was using geometric series formula, but it was not applicable here...I don't know where I have to start, and I don't know any hint about this series....  If anybody helps me, I would be very thankful to him.
4635
I was given this problem 30 years ago by a coworker, posted it 15 years ago to rec.puzzles, and got a solution from Barry Wolk, but have never seen it again.  Consider the series:  $$1, \frac{1}{2},\frac{\displaystyle\frac{1}{2}}{\displaystyle\frac{3}{4}},\frac{\displaystyle\frac{\displaystyle\frac{1}{2}}{\displaystyle\frac{3}{4}}}{\displaystyle\frac{\displaystyle\frac{5}{6}}{\displaystyle\frac{7}{8}}},\cdots$$  Each fraction keeps its large bars while being put atop a similar structure. This can also be represented as $$\frac{1\cdot 4 \cdot 6 \cdot 7 \cdot\cdots}{2 \cdot 3 \cdot 5 \cdot 8 \cdot\cdots}$$ terminating at $2^n$ for some $n$, where it is much closer to the limit than elsewhere.  The challenge:   Find the limit, not too hard by experiment In the last expression, find a simple, nonrecursive, expression to say whether $n$ is in the numerator or denominator Prove the limit is correct-this is the hard one.
101687
Find the value of $$S=\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{4}{2n+1}\right)$$    My Try:we have  $$S=2\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2n+1}\right)$$    $$S=2\left(1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\cdots\right)$$ so  $$S=2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\right)$$  But we know  $$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$  So  $$S=2(2-\ln 2)$$  Is this correct?
12639
$\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$  The general term seems to be $$T_r= \frac{r+2}{r(r+1)(r+3)}.$$  I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.  Do I have to use something else? Or am I missing some easy manipulation?
69759
I'm a PreCalculus student trying to find a rigorous proof that $\displaystyle\frac{1}{3} = 0.333\ldots$, but I couldn't find it. I think (just think) that this proof would start by proving that  $\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = \frac{1}{3}$. My guesses (assuming that proving that $\displaystyle\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i$ converges is trivial):  $\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = 3\cdot\sum_{i = 1}^{\infty}10^{-i} = 3\cdot\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i = 3\cdot\left(\frac{1}{1 - \frac{1}{10}}-1\right) = 3\cdot\left(\frac{10}{9}-1\right) = \frac{1}{3}$.  Questions: is this completely rigorous? Which flaws could be found in this proof? How can I improve it?  PS. I'm not sure how to tag this. Feel free to edit, if necessary.
45472
Given a fraction:  $$\frac{a}{b}$$  I now add a number $n$ to both numerator and denominator in the following fashion:  $$\frac{a+n}{b+n}$$  The basic property is that the second fraction is suppose to closer to $1$ than the first one. My question is how can we prove that?  What I have tried:  I know $\frac{n}{n} = 1$ so now adding numbers $a$ and $b$ to it would actually "move it away" from $1$. But I cannot understand why $\frac{a}{b}$ is actually farther away from $1$ than $\frac{a+n}{b+n}$.  Why is that? What does it mean to add a number to both the numerator and denominator?
115753
How can I go about finding the limit of $$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.
68737
$$\frac1{(1-x)^3} = \sum_{n=0}^\infty \binom{n+2}2x^n$$  I am not sure how to begin. $$\begin{aligned}\frac1{1-x}&amp;= \sum_{n=0}^{\infty} x^n\\\frac1{(1-x)^2}&amp;=\sum_{n=0}^{\infty} nx^{n-1}\end{aligned}$$ Then you would get: $$\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} nx^{2n - 1}$$ Which is nowhere near the same...?
64544
Can anybody explain what a fraction is in a way that makes sense. I will tell you what I find so confusing:   A fraction is just a number, but this number is written as a division problem between two numbers! There is no answer to this division problem, I mean if I try to calculate what the result is I'll just get a fraction as a result! So if I inputed this into my calculator: $\frac{1}{2}$, I will get a result of $0.5$, which itself is a fraction: $\frac{5}{10}$!
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 $\frac{\ln|x|}{2}+ C$?
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 $\displaystyle\frac{2x}{1}$?
76183
Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me.  I know that $0.999 \cdots$ (repeating) = $1$. So wouldn't $1 - \frac{1}{\infty} = 1$ as well? Because $\frac{1}{\infty} $ would be infinitely close to $0$, perhaps as $1^{-\infty}$?  So $1 - 1^{-\infty}$, or $\frac{1}{\infty}$ would be equivalent to $0.999 \cdots$? Or am I missing something? Is infinity something that can even be used in this sort of mathematics?
121291
I have a clueless friend who believes that   $$ \sum_{n=1}^\infty \frac{1}{2^n} $$  doesn't equal $1$ in the 'normal arithmetical sense'.   He doesn't believe that this series flat out equals $$1$$ Is he correct?
41068
Problem is to find sum  $$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$  What I knew doesn't apply on this problem  Some series are telescoping, some types are solvable using binomial , both look useless here Binomial gives  $$n (n-1) (n-2)x^3=1$$  $$n (n-1)( n-2)(n-3)x^4=4$$  What approach to use here?
39076
Show that $$\dfrac{3}{1\cdot2\cdot4}+\dfrac{4}{2\cdot3\cdot5}+\dfrac{5}{3\cdot4\cdot6}+\cdots+\dfrac{n+2}{n\cdot(n+1)\cdot(n+3)}\\=\dfrac{1}{6}\left[\dfrac{29}{6}-\dfrac{4}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3} \right]$$   Trial: I proceed in this way $$\dfrac{3}{1\cdot2\cdot4}+\dfrac{4}{2\cdot3\cdot5}+\dfrac{5}{3\cdot4\cdot6}+\cdots+\dfrac{n+2}{n\cdot(n+1)\cdot(n+3)}\\=\dfrac{1}{6} \left[\dfrac{18}{1\cdot2\cdot4}+\dfrac{24}{2\cdot3\cdot5}+\dfrac{30}{3\cdot4\cdot6}+\cdots+\dfrac{6n+12}{n\cdot(n+1)\cdot(n+3)} \right] $$Then stuck how to break $6n+12$ so that we get the following result.