Suppose I have something with length one unit. I divide it to two equal length $0.5$ unit and put left one in my left side and right one in my right side. I then do same for my right side and contact result's left one to my left side and replace my right side with it's right one.  Even if I do these infinitely, there's something at my right side remaining. So it seems that $$\frac12+\frac14+\frac18+\dots$$ never taste value $1$. Right?  NOT DUPLICATE: Actually  does not answer my question. because if I accept them then I have $$\frac12+\frac14+\frac18+\dots=\frac23+\frac29+\frac2{27}+\dots=1$$ Then I can define $$A_n = \frac12+\frac14+\frac18+\dots+\frac1{2^n}$$ And $$B_n = \frac23+\frac29+\frac2{27}+\dots+\frac2{3^n}$$ Then, it's quite can be seen that: $$A_n&lt;B_n , n\in\mathbb N$$  Then how could I accept that both series can reach each other at value $1$?!
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?
38461
I am trying to calculate the following series:  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$  and I managed to reduce it to this term  $$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$  And here I am stuck. I tried writing down a few partial sums but I can't see the pattern, $\frac{1}{2}-\frac{1}{2}+\frac{1}{5}+\frac{1}{4}-\frac{1}{3}+\frac{1}{8}+...$ I cant seem to find a closed formula that we can calculate for $S_n$  How should I go about solving this question
49817
Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
113546
Find the value of infinite series   $$ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{2}{n^2} \right) $$  I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer.
28609
I am sure there is something wrong i am doing but i just want someone to point it to me.  Why can't we say that $e^{\frac {\pi i}{3}} = \left(e^{\pi i}\right)^{\frac {1}{3}} = (-1)^{\frac {1}{3}} = -1$ Why do we have to calculate it by the formula as it is (which produces different value)  Another question is why can't we say that $ (-1)^{\frac {1}{2}} = \left(\left (-1\right)^{2}\right)^{\frac {1}{4}} = 1^{\frac {1}{4}} = 1$  Again i have to say that i know this is all wrong i just want to know why it is wrong
45377
I have been forewarned about it, I have read the  here, but I haven't seen a counter example where it doesn't work. I know that it isnt really a fraction, but does it effectively get the same result in all cases, or are there counterexamples I must be warned about.   I do not mean obvious counter examples, like $$\frac{dy}{dx}+\frac{du}{dv} = \frac{dydv+dxdu}{dx dv}$$ which as far as I know doesn't really mean anything.
109467
When answering , I came up with the following: Suppose we have a length $l$ and integrate $\frac 1 l$ over $l$: $$\int \frac {\operatorname d\!l}l=\ln l$$ Since the dimension of $l$ is length, and $\ln$ should have a dimensionless argument, this cannot be a 'good' integral in the physical sense, although it would be mathematically (omitting units). A way to solve this would be to reformulate the indefinite integral like this: for $c=-\ln (1 \text{meter})$. $$ \int \frac {\operatorname d\!l}l=\ln l+C=\ln l-\ln( 1\text{m})=\ln\left(\frac{l}{1\text{meter}}\right) $$ The question is whether the first integral given would be acceptable and how to deal with the two logarithms with bad arguments, which, when combined, aren't a problem anymore.
113458
Using the equivalency, I got the answer equal to $1$, however the answer is given as $\frac34$.    also gives $\frac34$.  Could someone tell me that the given answer is wrong or I made a mistake!
24413
So I am studying series for an exam right now and there is an example in the book I am studying (unfortunately the book is specific to my university so I cannot give any link) where certain series' sums are calculated. At some point the following ones are calculated:  $$\sum_{n=1}^{\infty}\left(\dfrac{1}{5}\right)^{n-1}=\frac54$$  and  $$\sum_{n=1}^{\infty} n\left(\dfrac{1}{5}\right)^{n-1}=\frac{25}{16}$$  I do not understand the reason that extra $n$ (the multiplier in the front) squares everything. Also I have no clue what to do when I have say   $$\sum_{n=1}^{\infty}n^2\left(\dfrac{1}{5}\right)^{n-1}$$  Do I cube everything? Is there a specific way to do it that I do not get? If there is some online paper, book chapter or whatever that could help me, please link me to it!
60322
Find the value of the following infinite series: $$\left\lfloor\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}\right\rfloor$$  Now, my doubt is whether it's​ $2$ or it's​ $3$. I'm not sure if it just converges to $3$ but not actually reaches it or if it completely attains the value of $3$.  I would to like to see the proof (formal) as well instead of just intuitions.
103850
I've looked up several related questions, but they do not answer what I am looking for.  Please give link if this is a duplicate.  What I eventually want to know is why  $$\lim_{n\to\infty}\left(1+\frac1n\right)^n$$  converges to a finite value.  I mean, we have to know that the limit converges before we can call it a constant, right?  I know that $f(n)=\left(1+\dfrac1n\right)^n$ is strictly increasing, but how do we know it is bounded above?  Also, I tried using the binomial theorem to write  $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left[\sum_{k=0}^n\frac{n!}{n^k k! (n-k)!}\right],\tag{1}$$  but I am unsure how to prove that the sum converges for $n\to\infty,$ because the general term depends on both $n$ and $k.$    Here is what I think will work.  Let $a_{n,k}=\dfrac{n!}{n^k k! (n-k)!}.$  Then $a_{n+1,k+1}=\dfrac{(n+1)!}{(n+1)^{k+1} (k+1)! (n-k)!}.$  Using the Ratio Test, we get  $$\lim_{n\to\infty\atop k\to\infty}\left|\frac{a_{n+1,k+1}}{a_{n,k}}\right|=\lim_{n\to\infty\atop k\to\infty}\left[\frac{1}{k+1}\left(\frac{n}{n+1}\right)^k\right].$$  where the limit on the RHS is less than 1.    That may take care of the case where $n\in\mathbb{N},$ so I suppose that (1) can be generalized using "."
115659
I'm trying to solve a limit problem but I've never encountered the one like this before.  $$ \lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)$$  I multiply the right side by $\frac{t\sqrt{1+t}}{t\sqrt{1+t}}$ and combine the terms to get:  $$ \lim_{t\to 0}\left(\frac{1 - t\sqrt{1+t}}{t\sqrt{1+t}}\right)$$  I can factor out the $t\sqrt{1+t}$ and I am left with:  $$1 - 1 = 0$$  This seems wrong to me. I can't explain why, maybe I just feel that getting a zero after all that work seems lame. Am I following the right steps?
114239
Find the value of $$ \lim_{n\to\infty} \sum_{k=n+1}^{2n} \frac{1}{k} $$ I think it become a series with each denominator approach infinity. so the value of the limit is 0. am i true ?
106805
So I just started learning series, and I am having trouble understanding them. I have a series below:  $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)2^n}$$  Can someone explain to me how I should go about this, from start to finish? I couldn't find any online websites that explained how to find sums, so if you have that, it would be greatly appreciated!
123584
Sorry for the oversimplified question, but does the series $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge? The ratio test of it gives the result of "1". Thanks a lot.
1574
Just encountered a problem in my BC Calculus sequences and series unit that I just can't figure out.   I don't know the latex, but the problem was to find compute  $$ \sum_{n=1}^\infty \frac{1}{2^nn} $$ The handout was just on convergence, so it was easy to see that by ratio test, it does converge, but I think I'm missing some ideas to be able to calculate the sum.   We have done Taylor series, I just don't quite know how to go backwards, although my guess is that you find a way to write the partial sums and find it's value as the limit approaches infinity.  How would you go about solving this?  Thanks!
40742
Find the limit of the following series:  $$ 1 - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} + \frac{1}{11} - \frac{1}{14} + \cdot \cdot \cdot  $$  If i go the integration way all is fine for a while but then things become pretty ugly. I'm trying to find out if there is some easier way to follow.