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12

Let $$\begin{align} S_1&=1-\frac15+\frac19-{1\over13}+\dots\\ S_2&=\frac13-\frac17+{1\over11}-{1\over15}+\dots \end{align}$$ so that the sum we seek is $S_1+S_2.$ T0 compute $S_1,$ consider $$f(x) = 1-{x^5\over5}+{x^9\over9}-{x^{13}\over13}+\dots$$ so that $$f'(x)=-x^4+x^8-x^{12}+\dots={-x^4\over1+x^4},\ |x|<1$$ and $$f(x)=\int_0^x{-t^4\over1+t^4}\...


10

Hint: write $\frac{1}{1+t^r}$ using the formula for a sum of a geometric series, and change the order of summation. Full solution: \begin{align} \lim_{t\rightarrow 1^-} (1-t) \sum_{r=1}^\infty \frac{t^r}{1+t^r} &= \lim_{t\rightarrow 1^-} (1-t) \sum_{r=1}^\infty t^r \sum_{n=0}^\infty (-t^r)^n = \\ &= \lim_{t\rightarrow 1^-} (1-t) \sum_{n=0}^\infty \...


5

Use the factorization: $$\dfrac{n}{4n^4+1} = \dfrac{n}{(2n^2-2n+1)(2n^2+2n+1)} = \frac 14\cdot\dfrac{1}{2n^2-2n+1}-\frac 14\cdot\dfrac{1}{2n^2+2n+1}$$


5

Hint: $$\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots=\sum_{k=1}^{\infty }\frac{1}{8k-7}+\frac{1}{8k-5}-\frac{1}{8k-3}-\frac{1}{8k-1}$$ $$=\sum_{k=1}^{\infty }(\frac{1}{8k-7}-\frac{1}{8k-1})+(\frac{1}{8k-5}-\frac{1}{8k-3})$$ $$=\sum_{k=1}^{\infty }(1+\frac{1}{8k+1}-\frac{1}{8k-1})+(\frac{1}{3}+\frac{1}{8k+3}-\frac{1}{8k-3})$$ $$=\frac{4}{3}+\sum_{...


5

We can write the last multiple sum as \begin{align*} \color{blue}{\sum_{i_1=1}^n\sum_{i_2=1}^{i_1}\sum_{i_3=1}^{i_2}i_3} &=\sum_{i_1=1}^n\sum_{i_2=1}^{i_1}\sum_{i_3=1}^{i_2}\sum_{i_4=1}^{i_3} 1\\ &=\sum_{1\leq i_4\leq i_3\leq i_2\leq i_1\leq n}1\tag{1}\\ &\,\,\color{blue}{=\binom{n+3}{4}}\tag{2} \end{align*} In (1) we ...


4

Hint: Use method of cancellation $$\frac {x^2} {1-x^4} = \left[\frac 1 {1-x^2} - \frac 1 {1-x^4}\right]$$ $$\frac {x^4} {1-x^8} = \left[\frac 1 {1-x^4} - \frac 1 {1-x^8}\right]$$


4

Let $f_k(n)$ be the closed form of the summation nested $k$ times. We know that $$f_1(n)=\frac12n(n+1)=\binom{n+1}{2}$$ $$f_k(n)=\sum_{j=1}^n f_{k-1}(j)$$ So for the next function $f_2(n)$ we have $$f_2(n)=\sum_{j=1}^n\binom{j+1}{2}=\sum_{j=2}^{n+1}\binom{j}{2}=\binom{n+2}{3}$$ By using the Hockey-stick identity (credits to Jean-Claude Arbaut). Similarly for ...


3

The easiest way is to construct the series from the original function explicitly $$ \cos u = \sum_{k=0}^\infty \frac{(-1)^n u^{2n}}{(2n)!} $$ In your case, you need to make $u^{2n}$ look like $\pi^{2n}/6^{2n} = (\pi/6)^{2n}$ so we set $u = \pi/6$ and the LHS becomes $$ \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}. $$


3

Of course, as commented, begin with the right side: $$\frac12(e^x-e^{-x})=\frac12\sum_{n=0}^\infty\frac{[1-(-1)^n]x^n}{n!}=(**)$$ But since $$1-(-1)^n=\begin{cases}0,&n\;\text{is even}\\{}\\2,&n\;\text{is odd}\end{cases}$$ We get: $$(**)=\frac12\sum_{n=0}^\infty\frac{2x^{2n+1}}{(2n+1)!}$$


2

By induction you can proof easily $$\sum\limits_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^k}} = \frac{1}{1-x^{2^n}}\sum\limits_{k=1}^{2^n-1}x^k$$ and with $\enspace\displaystyle \sum\limits_{k=1}^{2^n-1}x^k = \frac{x-x^{2^n}}{1-x}\enspace$ the formula is complete.


2

You've got a good notion! Integrating the partial sum $$1+2x+\cdots nx^{n-1}$$ gives you $$C+x+x^2+\cdots x^n,$$ for some constant $C,$ which is $$C-1+\frac{1-x^{n+1}}{1-x}.$$ Then, taking the derivative using the quotient rule gets you $$\begin{eqnarray}\frac{-(n+1)(1-x)x^n+1-x^{n+1}}{(1-x)^2} &=& \frac{-(n+1)x^n+(n+2)x^{n+1}+1-x^{n+1}}{(1-x)^2}\\ ...


2

A first transformation that gives an equivalent series which converges a it more rapidly is to break the series into even and odd terms, and let $n = 2k$ or $2k+1$. then you get $$ \sum_{k=1}^\infty \left( \frac{1}{\log (2k)} - \frac{1}{\log (2k+1)} \right) = \sum_{k=1}^\infty \frac{\log\left( 1+\frac 1{2k}\right)}{\log(2k)\log(2k+1)} $$ And now you can ...


2

We show the difference between the variant with general $y$ and $y=1$. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k}=\oint_{0<|z|<1} \frac{(1+z)^n}{z^{k+1}}\frac{dz}{2\pi i} \tag{1} \end{align*} We obtain \begin{align*} \color{...


2

Hint: Just a manipulation $$S(N,K)=\sum _{i=1}^N\sum _{j=i+1}^N\mathbb{I}(j-i\leq K)=\sum _{i=1}^N\sum _{j=i+1}^N\mathbb{I}(j\leq K+i)=\sum _{i=1}^N\sum _{j=i+1}^{\min \{K+i,N\}}1\qquad\qquad\qquad$$$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum _{i=1}^{N-K}\sum _{j=i+1}^{K+i}1+\sum _{i=N-K+1}^N\sum _{j=i+1}^N1.$$ Can you finish?


2

Split the sum on $i$ until $N-K$, and from $N-K+1$ to $N$ respectively, call them $S_1, S_2$, where we assume $1\le K < N$ as the other cases have been dealt with in the post. If $i\le N-K$, the sum on $j$ has all its $K$ terms, so $S_1=K(N-K)$ If $i> N-K$, the second sum has only $N-i$ terms, so $S_2=\sum_{i=N-K+1}^{i=N}{(N-i)}=\sum_{k=0}^{k=K-1}k=\...


2

Hint: Consider two cases: Case I is when $k = -k_0 + \alpha N$ for an integer $\alpha$. In this case $$e^{-2i\pi(k+k_0)/N} = ?$$ Case II is when $k \neq -k_0 + \alpha N$ for an integer $\alpha$. Use $$\sum_{n=0}^{N-1} q^n = \frac{1-q^{N}}{1-q}$$ to show that the sum is zero.


2

Whenever $k - N + k_0 = 0$ then our sum is $\sum_{n=0}^{N-1}e^{-2\pi i n} = N$. Otherwise, it's the sum of the roots of unity of order $N$ which is zero.


2

The easiest approach is to simply evaluate the inner sum directly. To this end, let $S(i)$ represent the inner summation; we consider cases where $i$ is even versus $i$ odd. Note that there are exactly $(i+2) - 3 + 1 = i$ terms in the sum. When the number of terms in the sum is even, they can be grouped into consecutive pairs. When the number of terms in ...


2

You can expand directly: $$S=\sum_{i=0}^{x-1} \Biggl(4+4\sum_{j=3}^{i+2} j(-1)^j\Biggr)=\sum_{i=0}^{x-1}4+4\sum_{i=\color{red}1}^{x-1}\sum_{j=3}^{i+2} j(-1)^j=\\ 4x+4\sum_{i=1}^{x-1}[-3+4-5+\cdots +(-1)^{i+2}(i+2)]=\\ 4x+4[\underbrace{(-3)}_{i=1\\ x=2}+\underbrace{(-3+4)}_{i=2\\ x=3}+\underbrace{(-3+4-5)}_{i=3\\ x=4}+\cdots +(-3+4-5+\cdots +(-1)^{x+1}(x+1))]=...


2

\begin{align}\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}&=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\\&=\frac12\left(2x+2\frac{x^3}{3!}+2\frac{x^5}{5!}+\cdots\right)\\&=\frac12\left(\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)-\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots\right)\right)\\&=\frac{e^x-e^{-x}}2.\end{align}


2

What you have is an iterated composition according to the terminology of Dixon J. Jones, A chronology of continued square roots and other continued compositions, through the year 2016 in contrast to a similar continued composition which would be a continued fraction in this case. The Wikipedia article Continued fraction has some indication of using products ...


2

Jacobi theta function ... $$ \vartheta_2(z,q) := 2 q^{1/4}\sum_{n=0}^\infty q^{n(n+1)}\cos((2n+1)z) $$ so that $$ \frac{\vartheta_1(0,q)}{2q^{1/4}} = \sum_{n=0}^\infty q^{n+n^2} $$ which is the function defined in the question title. The other answer: your function is not an elementary function.


1

$$p_n(x):=\sum_{i=1}^n x^i$$ is a polynomial, which you can differentiate term-wise, giving the polynomial $$p'_n(x):=\sum_{i=1}^n ix^{i-1}.$$ At the same time, $p(x)$ is the sum of terms of a geometric series, and for $x\ne1$, $$p_n(x)=\frac{x^{n+1}-1}{x-1}-1.$$ Then, for all $x\ne1$, $$p'_n(x)=\frac{(n+1)x^n}{x-1}-\frac{x^{n+1}-1}{(x-1)^2}.$$ The ...


1

The fraction is $+1$ or $-1$ depending on whether $i \lt \frac nm$ or $i \gt \frac nm$. If $m$ divides $n$ you have a term the divides by $0$ so the sum is undefined. Assuming you don't fall into the undefined trap, the summand is just $2$ as long as $i \lt \frac nm$, and $0$ after, so the sum is just $2\lfloor \frac nm \rfloor$. To convert it to an ...


1

In a strictly intuitive sense, the answer would be yes. You would probably have to develop some conventions for, say, "negative" squares and such - maybe draw them in a different color perhaps. And we would need a notion for reductions in the amount colored, which I guess could be analogized as erasing in some sense. If $\sum_n a_n$ converges to $S$, then ...


1

You could write it using a hypergeometric function: $$ \left( 1-y \right) ^{n}-{n\choose D+1} \left( -y \right) ^{D+1} {\mbox{$_2$F$_1$}(1,-n+D+1;\,D+2;\,y)} $$


1

$$ \begin{align} &\sum_{k=0}^\infty\left(\frac1{8k+1}+\frac1{8k+3}-\frac1{8k+5}-\frac1{8k+7}\right)\\ &=\sum_{k=0}^\infty\left(\frac1{8k+1}-\frac1{8k+7}\right)+\sum_{k=0}^\infty\left(\frac1{8k+3}-\frac1{8k+5}\right)\tag1\\ &=\sum_{k\in\mathbb{Z}}\frac1{8k+1}+\sum_{k\in\mathbb{Z}}\frac1{8k+3}\tag2\\ &=\frac18\sum_{k\in\mathbb{Z}}\frac1{k+\...


1

Here's a combinatorial way of thinking about it: first of all, note that we can go one level deeper and represent the innermost piece ($j$, or $k$, etc.) in your formulae as $\sum_{h=1}^j1$; this means that the formula start to look like $\displaystyle\sum_{m=1}^n1 =n$, $\displaystyle\sum_{m=1}^n\sum_{l=1}^m1=n(n+1)/2={n+1\choose 2}$, $\displaystyle\sum_{m=1}...


1

I do not think that there is a single value of $c$. So, let us consider that you look for the zero of function $$f(c_n)=\sum_{k=2}^{n}c_n^{\frac{1}{k\log k}}-n$$ which does not make much problems to solve numerically. For illustration purposes, let $n=10^k$; computing, we should obtain the following values $$\left( \begin{array}{cc} k & c_{10^k} \\ 1 &...


1

Let $k^3=k(k-1)(k-2)+ak(k-1)+ bk$ $\implies k=1,b=1$ $k=2,2^3=2a+2b\iff a=3$ Now for $k>2, $ $$k(k-1)(k-2)\binom nk=n(n-1)(n-2)\binom{n-3}{k-3}$$ Similarly $k\binom nk=?$ $k(k-1)\binom nk=?$ Finally use $(a+b)^m=\sum_{r=0}^n\binom mra^{m-r}b^r$ Set $a=b=1, m=n-3,n-2,n-1$


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