Articles of sequences and series

Closed form solution of Series contaning modified Bessel Function

The closed form solution of the series is $$\sum_{n=1}^\infty I_n(x) = \frac{(e^x-I_0(x))}{2}$$ What is the closed form solution of this one $$\sum_{n=1}^\infty \frac{I_n(x)}{n^2}$$ Of course, it is a convergent series.

If $A_n = \sum_{k = 1}^n \dfrac{k^6}{2^k}$, find $\lim_{n \to \infty} A_n$

Set $$A_n = \sum_{k = 1}^n \dfrac{k^6}{2^k}.$$ Find $\displaystyle \lim_{n \to \infty} A_n$. I tried solving this using a reduction method. That is, reducing the above series to an Arithmetico-Geometric series. We can see this by noting that $$\sum_{k = 1}^{\infty} \dfrac{k^m}{2^k} = \dfrac{2}{3} \left(1+\dfrac{2^m}{2} +\sum_{k=1}^{\infty} \dfrac{(k+1)^m-(k-1)^m}{2^k}\right)$$ which can be derived using infinite series. Using […]

How to find $\sum_{k=1}^n k^k$?

Actually question which I found: Find the sum of the series $1^1+2^2+3^3+ \cdots +n^n $ This question has been bothering me since a long time. Any help would be appreciated!

Simplification of $\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$

$$\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$$ The above sequence amounts to one of the following options: $\binom{100}{50}$ $\binom{100}{51}$ $\binom{50}{25}$ $\binom{50}{25}^2$

Infinite Product is converges

I am adding this problem since it is interesting and valuable to be verified here: Prove that the infinite product $\prod_{k=1}^{\infty}(1+u_k)$, wherein $u_k>0$, converges if $\sum_{k=1}^{\infty} u_k$ converges. What about the inverse problem? Thanks for any ideas.

Translations AND dilations of infinite series

Sometimes, when working with infinite series, it’s useful to add “dilated” or “translated” versions of the infinite series, term by term, back to the original. There are ways of making this rigorous for each. If you want to add a “translated” copy of a series to the original one, you can: attach the series coefficients […]

Closed form for this sum with hyperbolic cotangent $\sum _{n=1}^{\infty }\frac{\coth (xn)}{n^3}$

Is there a closed form for this sum? $$\sum _{n=1}^{\infty }\frac{\coth (xn)}{n^3}$$ I have tried using $$x\coth \left(xn\right)=\frac{1}{n}+2n\sum _{k=1}^{\infty }\frac{1}{n^2+\left(\frac{\pi }{x}\right)^2k^2}$$ To solve but got stuck at evaluating $$\sum _{n=1}^{ \infty}\sum _{k=1}^{\infty }\frac{1}{k^2\left(n^2+\left(\frac{\pi }{x}\right)^2k^2\right)}$$

closed form expression of a hypergeometric sum

After playing around with transforms of a certain parametric integral, I am inclined to think that the linear combination $$f(n):=\dfrac1{n-2}\left({\,}_2F_1(\dfrac{n-2}{4n},\dfrac12;\dfrac{5n-2}{4n};-1)\right)+\dfrac1{n+2}\left({\,}_2F_1(\dfrac{n+2}{4n},\dfrac12;\dfrac{5n+2}{4n};-1)\right)$$ has a closed form for integer $n$. I know for example that $f(3)=\dfrac{1}{12^{3/4}}\dfrac{\Gamma(\frac14)^2}{\sqrt{\pi}}$. Any ideas? Edit: putting $a:=\frac14-\frac1{2n}$, we can define $g(a):= \frac{8}{1-4a}f(\frac{8}{1-4a})$ to get arguments closer to the “standard” notation used in formula collections. […]

Why is $(1+\frac{3}{n})^{-1}=(1-\frac{3}{n}+\frac{9}{n^2}+o(\frac{1}{n^2}))$ and how to get around the Taylor expansion?

Let be $(u_n)$ a real sequence such that $u_0>0$ and that $\forall n \in \mathbb{R}$: $$\frac{u_{n+1}}{u_n}=\frac{n+1}{n+3}$$ Let be $(v_n)$ a real sequence such that $\forall n \in \mathbb{R}$: $$v_n=n^2u_n$$ Let’s determine the nature of $\sum\ln(\frac{v_{n+1}}{v_n})$ I did: \begin{align*} \frac{v_{n+1}}{v_n} &=\left(\frac{n+1}{n}\right)^2\frac{u_{n+1}}{u_n}\\ &=\left(1+\frac{1}{n}\right)^2\frac{n+1}{n+3}\\ &=\left(1+\frac{1}{n}\right)^3\left(1+\frac{3}{n}\right)^{-1}\\ \end{align*} and there I was stuck. A friend of mine gave me this […]

Prove $\lim\limits_{n \to \infty} \sqrt{X_n} = \sqrt{\lim\limits_{n \to \infty} X_n}$, where $\{X_n\}_{n=1}^\infty$ converges

Let $\{X_n\}_{n=1}^\infty$ be a convergence sequence such that $X_n \geq 0$ and $k \in \mathbb{N}$. Then $$ \lim_{n \to \infty} \sqrt[k]{X_n} = \sqrt[k]{\lim_{n \to \infty} X_n}. $$ Can someone help me figure out how to prove this?