I was given that $S=1+2r+3r^2+\cdots = \frac{1}{(1-r)^2}$ was an identity from my students, and I tried to prove directly using geometric series.  I got stuck and looked around online only to be told that taking the derivative of the series that converges to $\frac{1}{1-r}$.  I liked that proof because it was simple and easy, but unfortunately my student is in Alg II so I cannot explain her with derivatives.  My idea was to expand $S$ as $$(1+r+r^2+r^3+\cdots) +(r+r^2+r^3+\cdots)+(r^2+r^3+\cdots)+\cdots$$  but I get confused whether I should rewrite this as $$(1+r+r^2+r^3+\cdots) +r(1+r+r^2+\cdots)+r^2(1+r+\cdots)+\cdots$$  or  $$(1+r+r^2+r^3+\cdots) +((-1)+1+r+r^2+\cdots)+((-1-r)+1+r+\cdots)+\cdots$$  I was able to get the ID using the first way, but I am not sure if the step I took was valid due to some stuff I vaguely remember about manipulating an infinite series like that was not always allowed.  And on the second way I simply did not know where to go from there.  Can I have some help ?  It would also be helpful to know the criteria of when I can manipulate the order of how to add the series.
1602
How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to , but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.  In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
10195
If $a, b, c \in \mathbb{N}$, and $abc = 1$, prove that: $$S = \frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$$  Here is my try: $$\begin{align}\frac{2}{(a+1)^2+(b^2+1)} &amp; \le \frac{1}{\sqrt{(a+1)^2 \cdot (b^2+1)}}\\ &amp; = \frac{1}{(a+1) \cdot \sqrt{b^2+1}}\\ &amp; \le \frac{1}{(a+1) \cdot b}\\ &amp; \le \frac{1}{ab} \end{align}$$  So, $$S \le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{1}$$  Can you help me to continue it, please? Thanks!
44381
My problem is to find the sum of the series $$ S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots $$ where the terms are the reciprocals of the positive integers whose only prime factors are $2$s and $3$s.   I can see that every term in $S$ is of the form $\dfrac{1}{2^m3^n}$ where $m,n\geq 0$. I can also see that each term only occurs once, but I'm not really sure how to use this information effectively. Any ideas (apparently geometric series was a hint)?
10232
I came across a question where I needed to find the sum of the factorials of the first $n$ numbers. So I was wondering if there is any generic formula for this? Like there is a generic formula for the series: $$ 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} $$ or $$ 1^{2} + 2^{2} + 3^{2} + 4^{2} + \cdots + n^{2} = \frac{n(n+1)(2n + 1)}{6} $$  Is there is any formula for: $$ 1! +2! +3! + 4! + \cdots + n! $$ and $$ {1!}^2 +{2!}^2 +{3!}^2 + \cdots + {n!}^2 $$?  Thanks in advance. If not, is there any research on making this type of formula? Because I am interested.
51864
Consider, $$ S= \sum_{k=0}^{k=\infty} \frac{ k!}{x^{k}} (-1)^k$$ now this is, $$ S = ( 1 +D+D^2 +D^3...) ( \frac{1}{x})$$ using geometeric series $$ 1+D+D^2.. = \frac{1}{1-D}$$ So, $$ S= \frac{1}{1-D} \frac{1}{x}$$ $$ S = \frac{1}{x-1}$$ Therefore , for x&lt;1 $$ S= \sum_{k=0}^{k=\infty}(-1)^k \frac{ k!}{x^{k}}=\frac{1}{x-1}$$ Is this proof correct? Can I find an expression for l.h.s for x&gt;1 ?
53821
$R(t) \cdot R'(t) = 0$, which is what every source I can find tells me. Even though I understand the proof I don't understand the underlying concept. If $R(t)\cdot R'(t) = 0$, then $R'(t)$ is orthogonal to $R(t)$, right?   But you use the same derivative to find the tangent of a curve. Then somehow if you differentiate the tangent itself, you get the normal to the curve.  I really can't wrap my head around this. Could someone help me understand?
68698
Do you have any idea to prove the following identity via a combinatorial (or algebraic) method?  $\sum_{n=r}^{n=\infty} {n \choose r}^{-1} = \frac{r}{r-1}$  This is  of the book Chen C.C., Koh K.M. Principles and techniques in combinatorics. The book does not give a solution, although it mentions: "see H. W. Gould, Combinatorial Identities, Morgantown, W.V. (1972), 18-19".  many thanks in advance, Shahram
50695
I have to prove that  $$\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)\cdots (k+p)}=\dfrac{1}{p!p}$$  How can I do that?
3452
Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\cdots\left(1+\dfrac{1}{n^3}\right)&lt;3$   for all positive integers $n$   This problem is copied from Math Olympiad Treasures by Titu Andreescu and Bogdan Enescu.They start by stating that induction wouldn't directly work here since the right hand side stays constant while the left increases.They get rid of this problem by strengthening the hypothesis.$$\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\cdots\left(1+\dfrac{1}{n^3}\right)\le3-\dfrac{1}{n}$$ and then proceed by induction.  The problem is that I can't find a motivation for the above change.I mean,we could have subtracted a lot of things from the RHS but what should nudge us to try $\dfrac{1}{n}$?The rest of the proof is quite standard,but I can't see how I am supposed to have thought of it.Is it just experience?Or is it a standard technique?A little guidance and motivation will be appreciated.
76607
Here is a famous problem posed by Ramanujan     Show that $$\left(1 + \frac{1}{1\cdot 3} + \frac{1}{1\cdot 3\cdot 5} + \cdots\right) + \left(\cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}\right) = \sqrt{\frac{\pi e}{2}}$$   The first series seems vaguely familiar if we consider the function $$f(x) = x + \frac{x^{3}}{1\cdot 3} + \frac{x^{5}}{1\cdot 3\cdot 5} + \cdots$$ and note that $$f'(x) = 1 + xf(x)$$ so that $y = f(x)$ satisfies the differential equation $$\frac{dy}{dx} - xy = 1, y(0) = 0$$ The integrating factor here comes to be $e^{-x^{2}/2}$ so that $$ye^{-x^{2}/2} = \int_{0}^{x}e^{-t^{2}/2}\,dt$$ and hence $$f(x) = e^{x^{2}/2}\int_{0}^{x}e^{-t^{2}/2}\,dt$$ Thus the sum of the first series is $$f(1) = \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt$$ But I have no idea about the continued fraction and still more I am not able to figure out how it would simplify to $\sqrt{\pi e/2}$ at the end.  Please provide any hints or suggestions.  Update: We have $$\begin{aligned}f(1) &amp;= \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt = \sqrt{e}\int_{0}^{\infty}e^{-t^{2}/2}\,dt - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\\ &amp;= \sqrt{\frac{\pi e}{2}} - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\end{aligned}$$ and hence we finally need to establish $$\sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt = \cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}$$ On further searching in Ramanujan's Collected Papers I found the following formula $$\int_{0}^{a}e^{-x^{2}}\,dx = \frac{\sqrt{\pi}}{2} - \cfrac{e^{-a^{2}}}{2a+}\cfrac{1}{a+}\cfrac{2}{2a+}\cfrac{3}{a+}\cfrac{4}{2a+\cdots}$$ and it seems helpful here. But unfortunately proving this formula seems to be another big challenge for me.
70280
Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$   $\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$  So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$  So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$  $$  = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$  $$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$  So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$  Can we prove it Using Combinatorial way,bcz above method is very lengthy  If yes then plz explain here  Thanks
19547
Prove by Mathematical Induction . . .  $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$  I tried solving it, but I got stuck near the end . . .  a. Basis Step:      $(1)(1!) = (1+1)!-1$      $1 = (2\cdot1)-1$      $1 = 1 \checkmark$   b. Inductive Hypothesis     $1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$   Prove k+1 is true.     $1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$      $\big[RHS\big]$      $(k+2)!-1 = (k+2)(k+1)k!-1$      $\big[LHS\big]$      $=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)      $= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)      $= (k+1)!-1 + (k+1)(k+1)k!$      $= (k+1)k!-1 + (k+1)^{2}k!$   My [LHS] looks nothing like my [RHS] did I do something wrong?  EDIT:     $ = (k+1)k! + (k+1)^2k! -1 $      $ = (k+1)(k!)(1 + (k+1))-1$      $ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$
117249
If $f(s) = (1+s)^{(1+s)^{(1+s)/s}/s}$, show that $\lim_{s \to \infty} f(s)/s = 1$.  This function comes up in the parameterization of the solutions to $x^y = y^x$. See for example, here:   If we write $y = rx$, we get $x = r^{1/(r-1)}$ and $y=r^{r/(r-1)}$.  Then $x^y =(r^{1/(r-1)})^{r^{r/(r-1)}} =r^{r^{r/(r-1)}/(r-1)} $.  Finally, if we write $r = 1+s$, this is $f(s) = (1+s)^{(1+s)^{1+1/s}/s} $.  Wolfy says that, around $s=0$, $$f(s) =e^e +  \dfrac{e^{1 + e} s^2}{24}  - \dfrac{e^{1 + e} s^3}{24}  + \dfrac{e^{1 + e} (219 + 5 e) s^4}{5760} + O(s^5) $$ and, around $s = \infty$, $$f(s) =s + (\log^2(1/s) - \log(1/s) + 1)  + \dfrac{\log^4(1/s) - 3 \log^3(1/s) + 5 \log^2(1/s) - 6 \log(1/s) + 2}{2 s} + O((1/s)^2), $$ calling this a generalized Puiseux series.  My question is: How to show that that expansion at $s = \infty$ is correct. I would be satisfied with a proof that, as the title says, $\lim_{s \to \infty} \dfrac{f(s)}{s} = 1 $.  It might help to note that  $\dfrac{y}{x} = r$ and $\dfrac{r}{s} = \dfrac{1+s}{s} = 1+\dfrac1{s} $.
64081
I have to show that the double serie $$\sum_{k=1}^{\infty} \sum_{i=1}^{\infty} ((\frac{1}{k+1} \cdot (\frac{k}{k+1})^{i}) - (\frac{1}{k+2} \cdot (\frac{k+1}{k+2})^{i}))$$ convergence to $-\frac{1}{2}$. Wolfram Alpha also says this series convergence to $-\frac{1}{2}$. Unfortunately, I don't know the calculation route.
17401
So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} &lt; 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither.
38743
I'm in trouble with this homework.  Find the sum of the series $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$, where the terms are the inverse of the positive integers whose only prime factors are 2 and 3.  Hint: write the series as a product of two geometric series.  Ok, I found some patterns in these numbers, but I can't find the two series.
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.