Articles of sequences and series

Find the exact value of the infinite sum $\sum_{n=1}^\infty \big\{\mathrm{e}-\big(1+\frac1n\big)^{n}\big\}$

How can we find the exact value of the infinite sum $$ \displaystyle\sum_{n=1}^\infty \left\{\mathrm{e}-\Big(1+\frac1n\Big)^n\right\}? $$ This problem appears in: T. Andreescu, T. Radulescu & V. Radulescu, Problems in Real Analysis: Advanced Calculus on the real line, p.114.

Finding relatives of the series $\varphi =\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}$.

Consider $\varphi=\frac{1+\sqrt{5}}{2}$, the golden ratio. Bellow are series $(3)$ and $(6)$ that represent $\varphi$ $$ \begin{align*} \varphi &=\frac{1}{1}+\sum_{k=0}^{\infty}\cdots&(1)\\ \varphi &=\frac{2}{1}+\sum_{k=0}^{\infty}\cdots&(2)\\ \varphi &=\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}&(3)\\ \varphi &=\frac{5}{3}+\sum_{k=0}^{\infty}\cdots&(4)\\ \varphi &=\frac{8}{5}+\sum_{k=0}^{\infty}\cdots&(5)\\ \varphi &=\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(2(k+1))!}{((k+1)+1)!(k+1)!2^{4(k+1)+3}}&(6)\\ \vdots&\\ \end{align*} $$ When looking at the leading terms of $(3)$ and $(6)$ $\;\frac{3}{2}$ and $\frac{13}{8}$ respectively, one is tempted to conjecture that there are similar formulas […]

Prove that $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ converges and find its sum

Since this is not a geometric series, I know that I should use the definition of a convergent series, so $$S_n = \sum_{i=1}^n \ln\left(\frac{i(i+2)}{(i+1)^2}\right)$$ After this, I tried two different ways: 1) I simplified the fraction to read $$\frac{\ln(i^2+2i)}{i^2+2i+1}$$ and then I used long division to get $$\ln((1)-\frac{1}{(i+1)^2})$$ However, once I start plugging in i […]

Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} – \frac{1}{2 (n+1) (n+2)}$

Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} – \frac{1}{2 (n+1) (n+2)}$. I tried using the partial fraction decomposition $a_j = \frac{1}{2j} – \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don’t see how that helps.

Simplify $\left({\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}\right)\left({\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}\right)^{-1}$

Simplify $$\frac{\displaystyle\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}{\displaystyle\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}$$ I don’t have any good idea. I need your help.

Sum $\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$

$\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}=\sum_{n=i}^{\infty} \frac {1}{{2n \choose n-i}}$ is a very interesting one. Here is what I have from WolframAlpha. $\displaystyle \sum_{n=0}^{\infty} {2n \choose n}^{-1}=\frac{2}{27}(18+\sqrt{3}\pi)$ $\displaystyle \sum_{n=1}^{\infty} {2n \choose n}^{-1}=\frac{1}{27}(9+2\sqrt{3}\pi)$ $\displaystyle \sum_{n=1}^{\infty} {2n \choose n-1}^{-1}=\frac{1}{27}(9+5\sqrt{3}\pi)$ For $i \geq 2$, WA just comes up with closed forms involving generalised hypergeometric functions. Here is one example. […]

Evaluate the series $\sum_{n=1}^{\infty} \frac{2^{}+2^{-}}{2^n}$

How can I obtain the limit of the series $$\sum_{n=1}^{\infty} \frac{2^{[\sqrt{n}]}+2^{-[\sqrt{n}]}}{2^n}$$ Where $[\ \ ]$ is Nearest Integer Function.

Geometric series of matrices

I am currently reading ‘Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach’ by J. Hubbard and B. Hubbard. In the first chapter, there is the proposition: Let A be a square matrix. If $|A|<1$, the series $$S=I+A+A^2+\cdots$$ converges to $(I-A)^{-1}$. The proof first shows that $$S_k(I-A)=I-A^{k+1}$$ and similarly $$(I-A)S_k=I-A^{k+1}$$ where $S_k$ is the […]

Questions concerning the Integration of Integer Tetration

I’ve been interested in finding the antiderivative of integer tetration, a function defined as iterative exponentiation. Integer tetration is written as $^n$$x$ where $^1$$x =x$, $^2$$x =x^x$, $^3$$x =$ $x^{\scriptscriptstyle x^{x}}$ and so forth where $n=1,2,3\ldots$ (Further info can be found on Wikipedia. One solution to the problem is found on the MathWorld page (see […]

Infinite product: $\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}$

This question already has an answer here: How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$ 3 answers