I'm working on Project Euler, :     The series, 11 + 22 + 33 + ... + 1010 = 10405071317.      Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.   This would be the equivalent of: $$ f(n) = \sum_{i = 1} ^ n i^i $$  The original question aside, it got me thinking. The rate of change from n to n+1 would be i^i, so if I could take the integral of ii, I could get a generalized formula for any given n. Upon plugging it into Wolfram Alpha, I got a monstrous expansion, but it said, "no result found in terms of standard mathematical functions". It's been years since I've taken calculus and I'm mostly a computer science guy to begin with, but is this premise sound? And if not, where did I go wrong?
13823
I'm trying to find $\int x^x \, dx$, but the only thing I know how to do is this:  Let $u=x^x$.  $$\begin{align} \int x^x \, dx&amp;=\int u \, du\\[6pt] &amp;=\frac{u^2}{2}\\[6pt] &amp;=\dfrac{\left(x^x\right)^2}{2}\\[6pt] &amp;=\frac{x^{2x}}{2} \end{align}$$  But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong.
17609
I am trying to find the smallest number if we are given the sum of its digits. Suppose that sum of digits  is 9 then it should be 9 instead of 18,36,63 and similarly if sum of digits is  11 then desired answer is 29 not 92 or any other number bigger than 29.I tried to write sum of all numbers upto 53 and got this but i am not able to come up with a general formula.                         from 1 to 9 it is just 9.FROM 10 to 18 it is 19,29,39,...... and for 19(1+9) it is 199  that is increase by 100. from 19 to 27 it is 199,299,399,499.... and for 28 it is 1999 that is increase by 1000. from 29 to 36 it is 29999,39999,49999,..... for 37 it is 19999 (increse by 10000). for 38 to 45 it is it is 29999,39999,49999,........999999 and for 46 it is 199999 (increased by 100000)
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: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
59684
I have encountered this question in a coding competition and want the formula for $f(n)$. Is there any way to do this?
10912
Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $\pi/4, \log 2$ and similar alternating series etc.  One series that popped into our discussion was $\sum\limits_{n=1}^{\infty} \frac{1}{n^n}$.  Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $\approx 1.29129$.  I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).  We were joking that it should have something to do with $\pi,e,\phi,\gamma$ or at the least it must be a transcendental number :-).  My questions are:   What does this series converge to? Does this series arise in any context and are there interesting trivia to be known about this series?   I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results).
69428
Does anyone know what is the sum function $f(x)$ of the series $\displaystyle\sum_{n=1}^\infty \frac{\cos(nx)}{n^2}$? I have no idea how to find a sum function... Any help would be appreciated.
60398
I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?  I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$  So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$?   With the binomial theorem, I get it equal to $$ \sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}). $$ Can this be made nicer? Thanks.
121100
While Solving some integral problem, I encountered the following infinite series:  $$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$  I have tried many methods including partial fractions... I seek help! Please provide hints if you don't have the complete answer.
66751
Actually not sure how to approach this... but I may be missing something:  Replacing the sequence:  $x_{n}=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}-2\sqrt{n},\,\,\,\, n=1,2,....$  By the corresponding series, invesigate it's convergence.  Hint: Take $a_{1}=x_{1}$ and $a_{n}=x_{n}-x_{n-1}$ for $n&gt;1$. Then $x_{n}$ is a sequence of partial sums for $\sum_{k=1}^{\infty}a_{k}$. Use an expicit formula for $a_{n}$ and use the comparison in limit test to show that the series converges.  Note: This looks easy, but as I said... there's something not quite connecting. When writing out the summation, I'm not entirely sure if they're compatible, as one would go from k=1 to n, and the other would go from k=1 to n-1, yet still vary with k... so do I need to use another pronumeral for it? Or can I simply use k-1 instead of k?
117844
I am wondering how to evaluate the following sum:  $$\sum_{n=0}^{\infty}\frac{1}{(n^4+n^2+1)n!}.$$  In wolfram alpha I find it is equal to $e/2$ .  I have used the residue method but I didn't succeed and also using digamma function is still hard for me, my problem is treating the $n!$.
93886
How can I find the value of the following sum? $$\sum_{n=0}^{\infty}\arctan(\frac{1}{F_{2n+1}})$$ $F_n$ is the Fibonacci number.($F_1=F_2=1$)
103880
Just checking. $2^n$ ($n \to \infty$) tends to $\infty$. + $3^n$ also ($n \to \infty$) tends to $\infty$  so the sum gets me $\infty$.  Now $(\infty)^{1/\infty}$ : $(\infty)^0 = 1$  I see no other way.  Theorem: $n^{\frac{1}{n}} = 1$ as $n \to \infty$.  Analogous:  $(2^n - 3^n)^\frac{1}{n}$ = $(\infty  - \infty)^0$ = 1 ???  Is it ok to suppose infinity on any integer $k^n$ as $n$ goes to infinity ?
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.  :|
13007
What is the sum of $$\sum_{n=0}^{\infty}\frac{1}{2^n(n+1)}$$    I've spent an insane amount of time on this problem. I checked on Wolfram and it gives $ln(4)$, which I assume you get from $2\ln(2)$.   I know that the integral of $\frac{1}{n+1}$ is $\ln(n+1)$. What am I missing?  Thank you!
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.
87270
This comes from the comments section of this question , credits .  The statement is  $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)=1$$  This looks really interesting, so I was wondering if anyone has any ideas/suggestions or even better the whole proof?  Disclaimer: I added that link specifically because this might look like homework, leading to "What have you tried?". So the disclaimer is, I actually haven't tried anything yet, because yes, asking the community is just easier. It would be great if someone could answer this though since I, personally at least, think that this equation is very interesting and adds value to any math enthusiast stumbling upon it here someday. :) I will myself be trying it meanwhile too though(it really is interesting) , and add and accept my solution if I am successful and there are no answers yet.     I tried  for $\sum_{k=0}^n\frac1{k!}$ and it yields  $$\sum_{k=0}^n\frac1{k!} = \frac{e\Gamma (n+1,1)}{\Gamma(n+1)}$$  where $\Gamma(a,r)$ is the incomplete gamma function and $\Gamma(a)$ is the euler gamma function. . Might be of help, might not. Just thought it'd be worth adding.
55477
I recently, out of interest, tried to prove Euler's formula $\zeta{(2n)}=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}$ for all $n\in\mathbb{N}$. I adapted  Euler's original proof for $\zeta(2)=\frac{\pi^2}{6}$:  We have the well known formulas $\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$ and $\sin(x)=x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)$. The product formula can be transformed into a series expansion as well but in order to keep it neat, some definitions first:  For the real sequence $\left(a_i\right)_{i\in\mathbb N}$ with $\sum a_i$ convergent, define $$ A_n:=\sum_{i=1}^{\infty} a_i^n $$ $$ \alpha_n:=\sum_{(x_1,\dots,x_n)\in M_n}\prod_{r=1}^{n} a_{x_r} $$ Where $M_n:=\{(x_1,\dots,x_n)\in\mathbb N\space|\space x_1&lt;\dots&lt;x_n\}$. Now we write $c_i=-\frac{1}{i^2\pi^2}$. When we expand the product, we get only odd exponents of $x$ and the coefficient of $x^{2k+1}$ for $k≥1$ is $\gamma_k$. By comparison of the coefficients, we obtain $\gamma_k=\frac{(-1)^k}{(2k+1)!}$. Now, through telescoping, it can be shown algebraically that for every sequence $\left(a_i\right)_{i\in\mathbb N}$ with $\sum a_i$ convergent we have: $$\sum_{i=1}^{n}{(-1)^{n-i}\alpha_{n+1-i}\space A_{i}}=A_{n+1}+(-1)^{n+1}(n+1)\alpha_{n+1}$$ For all $n\in\mathbb N_0$. Since $C_n=\sum_{i=1}^{\infty} \left(-\frac{1}{i^2\pi^2}\right)^n=\frac{(-1)^n}{\pi^{2n}}\zeta{(2n)}$ we have: $$ \sum_{i=1}^{n}{(-1)^{n-i}\gamma_{n+1-i}\space C_{i}}=\sum_{i=1}^{n}{(-1)^{n-i}\frac{(-1)^{n+1-i}}{(2n-2i+3)!}\frac{(-1)^i}{\pi^{2i}}\zeta{(2i)}}=\sum_{i=1}^{n}{(-1)^{i-1}\frac{\zeta{(2i)}}{(2n-2i+3)!\pi^{2i}}}=\frac{(-1)^{n+1}}{\pi^{2n+2}}\zeta{(2n+2)}+(-1)^{n+1}(n+1)\frac{(-1)^{n+1}}{(2n+3)!}=\frac{(-1)^{n+1}}{\pi^{2n+2}}\zeta{(2n+2)}+\frac{(n+1)}{(2n+3)!} \iff \sum_{i=1}^{n+1}{(-1)^{i-1}\frac{\zeta{(2i)}}{(2n-2i+3)!\pi^{2i}}}=\frac{(n+1)}{(2n+3)!} $$ For all $n\in\mathbb N_0$. Now we are ready to prove the formula $\zeta{(2n)}=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}$ with induction. With the equation above we can easily verify that it is true for $n=1$; therefore we assume that it holds for all $1≤i≤n$. The assertion $\zeta{(2n+2)}=(-1)^{n}\frac{(2\pi)^{2n+2}}{2(2n+2)!}B_{2n+2}$ is then equivalent to: $$ \sum_{i=1}^{n+1}{(-1)^{i-1}\frac{(-1)^{i-1}\frac{(2\pi)^{2i}}{2(2i)!}B_{2i}}{(2n-2i+3)!\pi^{2i}}}=\frac{(n+1)}{(2n+3)!}\iff \sum_{i=1}^{n+1}{\binom{2n+3}{2i}2^{2i-1}B_{2i}}=n+1\iff \sum_{i=0}^{2n+2} \binom{2n+3}{i}2^iB_i=0 $$ This identity can be proven by comparing the coefficients in the equation $\frac{e^x-1}{x}\frac{2x}{e^{2x}-1}=\frac{2}{e^x+1}$.   Is this proof valid? Nonetheless, if it is, it isn't very elegant. Are there faster ways to prove it?