$$S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12}\ldots\text{(to infinity)}$$  Rearranged, this series looks like: $$S = \left(1 - \frac12\right) - \left(\frac14\right) + \left(\frac13 - \frac16\right) - \left(\frac18\right) + \left(\frac15 - \frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac17 - \frac1{14}\right) \ldots\text{(to infinity)}\\  S = \left(\frac12\right) - \left(\frac14\right) + \left(\frac16\right) - \left(\frac18\right) + \left(\frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac1{14}\right) \ldots\text{(to infinity)}$$  This rearranged infinite series contains every number that the original infinite series had.  Further,  $$ 2S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12} \ldots\text{(to infinity)}$$ Thus: $2S = S$ $2 = 1$ Mathematics disproven.  Sorry.  Jokes aside, I know that infinite series can be calculated in different ways to get different results.  My question is: Why?  While it makes sense with Grandi's and similar series, it doesn't make sense to me for a series whose final term is $\frac1\infty = 0$.
123617
I don't understand that why the terms of an absolutely convergent series can be rearranged in any order and all such rearranged series converge to the same sum.  my textbook gives me an example that it is not so for conditionally convergent series  $1-\frac12+\frac13-\frac14+\ldots=\ln2$ $1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16...=\frac32\ln2$   which I also don't know how to prove
95733
Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$  I use software to complete the  series is  $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$  I have no idea about it.  :|
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!
2819
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$   I found that:  $$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$  and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
5055
I was messing around with some math formulas today and came up with a result that I found pretty neat, and I would appreciate it if anyone could explain it to me.   The formula for an infinite arithmetic sum is $$\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},$$ so if you want to find the sum of the natural numbers from $1$ to $n$, this equation becomes $$\frac{n^2+n}{2},$$ and the roots of this quadratic are at $n=-1$ and $0$. What I find really interesting is that $$\int_{-1}^0 \frac{n^2+n}{2}dn=-\frac{1}{12}$$ There are a lot of people who claim that the sum of all natural numbers is $-\frac{1}{12}$, so I was wondering if this result is a complete coincidence or if there's something else to glean from it.
51538
So I have this problem for midterm reviews:  $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=\text{ ?}$$  I know that you can find the series form of a natural log, as shown here:  $$\ln\left(1-\frac{1}{n^2}\right)=-\sum_{k=2}^\infty \left(\frac{1}{n^{4k}}\right)\left(\frac{1}{2k}\right) $$  But the above doesn't seem to help very much since it results in two summation notations mushed together. Is there a somewhat nontedious way to go about this? Thanks! All help appreciated.
124106
I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if there is  a better way to solve it. Here's how I did it:  First I decomposed the series into the following sums:  $S_1  = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$  $S_2 = \frac{7}{64} + \frac{7}{512} + \dots = \frac{\frac{7}{64}}{\frac{7}{8}}$  $S_3 = \frac{19}{512} + \frac{19}{4096} + \dots = \frac{\frac{19}{512}}{\frac{7}{8}}$  And deduced that the sum can be written as $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{3n^2 - 3n + 1}{8^n}$  $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ is easy to evaluate -- it's $\frac{1}{7} $by geometric series  $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ can be evaluated in a whole host of ways to get an answer of $\frac{8}{49}$.  It remains to evaluate $\displaystyle \sum_{n = 1}^{\infty} \frac{n^2}{8^n}$, for which I took a similar approach as the cubics by decomposing it into many sums:  $T_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$  $T_2 = \frac{3}{64} + \frac{3}{512} + \dots = \frac{\frac{3}{64}}{\frac{7}{8}}$  And so forth, coming to the conclusion that it is equal to $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{2n-1}{8^n}$  Now, I used this information and the above values for $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ and $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ to get the sum as $\frac{776}{2401}$, which is confirmed by WA.  So, I would like to reiterate here: Is there a simpler way to compute this sum, and are there any known generalizations for this problem given an arbitrary $a$ in the denominator and arbitrary $k$ as the exponent in the numerator?
38858
Prove that:     $$1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$   I know only this method:   $\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+....=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-...$  But, unfortunately, I could not a hint.
101295
I need to study this sum: $\sum_{n=1}^{\infty} \frac{n!\, i^n}{n^n}$. Taking: $$\lim_{n\rightarrow \infty} \; \left|\left(\frac{n!\: i^n}{n^n}\right)^{\frac{1}{2}}\right|$$ $$\rightarrow \lim_{n\rightarrow \infty}  \left|\left(\frac{n!}{n^n}\right)^{\frac{1}{2}}\right|$$  Using Stirling approximation considering the limit:  $$\ln n! \approx n \ln\:n-n$$ then $n! \approx (\frac{n}{e})^n$ (is this correct?):  $$\rightarrow \lim_{n\rightarrow \infty} \left|\frac{n}{n\, e}\right| = \frac{1}{e}$$ This doesn't make too much sense because I know the series diverges.  I think I'm missing a $\sqrt{2\pi n}$ in Stirling approximation but I don't understand why that pops out taking the $e^{(\;)}$ from the first expression.
123432
I have the following series  $$\sum_{k=1}^\infty \frac{(k-1)!}{(k+N)!},\quad\text{where }N \in \mathbb{N}. $$  I have found out that the series is equal to  $$\sum_{k=1}^\infty \biggl(\frac{1}{k}\cdot \frac{1}{k+1}\cdot _{...} \cdot \frac{1}{k+N}\biggr)$$  I also know that for $N=1$ we can use partial fraction expansion and we get a telescoping sum of $1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + ... + \frac{1}{k+1}$ which leaves $1$ if $k$ is heading to $\infty$.  Does anyone have any idea how I can go about doing this ?
122723
I'm noticing some things:  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/3}-\frac{3}{2}n^{2/3} \right)=\zeta(1/3)$$  Note $\int n^{-1/3} dn=\frac{3}{2}n^{2/3}+c$  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/2}-\frac{2}{1}n^{1/2} \right)=\zeta(1/2)$$  And  $$\int n^{-1/2}dn=2n^{1/2}+c$$  It seems as though  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/s}-\frac{s}{s-1}n^{(s-1)/s} \right)=\zeta(1/s)$$  If $s \neq 1$, may someone please explain why.
23735
Show that $\frac{1}{s-1}+\sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} - \frac{1}{n^{s-1}}\right)\right]$ converges when $0&lt;s&lt;1$.  Note that this series is equivalent to $\sum_{n=1}^{\infty}\frac{1}{n^s}$. It might seem counterintuitive that the series converges if $0&lt;s&lt;1$.
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!
31037
An  which I am not able to solve.  Let $\mathsf{S_{n}}$ denote the $n$-th partial sum of the series: $$\sum\limits_{r=1}^{\infty} \frac{1}{r(r+1)}$$  Prove that for every integer $k&gt;1$, there exists integers $m$ and $n$, such that $$s_{m}-s_{n} =\frac{1}{k}$$
22985
The series is given by ($a&gt;0$) $$\frac{1}{a} +\frac{1}{a+1} - \frac{1}{a+2} +\frac{1}{a+3} +\frac{1}{a+4} -\frac{1}{a+5} +\cdots$$ So how can I move forward? I  can not find the general term  of the series. If someone can give the general term then I can go further.
87603
This is the Grandi's series: $1-1+1-1+1-1+\dots$  The series can be equal to $0$   $$(1-1)+(1-1)+(1-1)+\dots=0+0+0+\dots=0,$$  or to $1$   $$1-(1-1)-(1-1)-(1-1)-\dots=1-0-0-0\dots=1,$$  or to $1/2$   $$S=1-1+1-1+1-\dots,\quad\quad    S=1-(1-1+1-1+1-1+1-...)$$ $$\Rightarrow   S=1-S\Rightarrow     2S=1\Rightarrow    S=1/2$$  Isn't this a contradiction? The integers are closed under addition and subtraction, but we get a fraction. Why?
47236
My friend's trivia league had this math question:  $$\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right]$$  After computing a few values, one could guess the answer is $e$ = 2.718...But how can we prove that is the limit?  Someone offered up a hand-wavy proof like this:  \begin{align} \lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right] &amp; = \lim_{n \to \infty} \left[\frac{(n+1)(n+1)^{n}}{n^n} - \frac{n \cdot n^{n-1}}{(n-1)^{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\frac{(n+1)^{n}}{n^n} - n\frac{n^{n-1}}{(n-1)^{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(\frac{n - 1 + 1}{n-1} \right)^{{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(1 + \frac{1}{n-1} \right)^{n-1} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)e - n \cdot e \right] \\ &amp;= \lim_{n \to \infty} e \\ &amp;= e \end{align}  The part about substituting $e$ is hand-wavy since technically this is an indeterminate form of $\infty - \infty$. And using $e$ as as upper bound did not lead me to an easy proof either.  Is there a way to rigorously prove the limit? I tried a few approaches: (a) sandwiching the limit--I could prove $e$ was a lower bound, but I could not find a suitable upper bound converging to $e$, (b) using L'Hopital's rule with no luck, (c) using the mean value theorem--but that also got complicated.  So this is a pretty tough problem to ask at trivia! Is there a way to prove this limit formally?  Sources  Trivia question:   Thread on trivia:   Hand-wavy proof:   Idea for mean value theorem: