Finding a closed formula for a sum

Question

I wrote this sum (out of the blue) and wondered if it has a closed form:

$$\sum_{k=1}^{\infty} L^{\frac{1}{k}} \cdot(-1)^{k+1}$$ where $L \in \mathbb{N}$

I thought of a sum that would use "$\text{k-root}$" but with alternating sign ($+$ to $-$ etc..)

I couldn't find a way to do so, the only thing I did is write a program that calculates it, so I post here for help. Thank you so much!! :-)

Answer 1

It doesn’t converge unless $L=0$. For $n\ge 1$ let $s_n=\sum_{k=1}^nL^{1/k}(-1)^{k+1}$. If $L\ge 1$, then $L^{1/k}\ge 1$ for all $k\ge 1$, so $|s_{n+1}-s_n|\ge 1$ for all $n\ge 1$. Thus, the sequence of partial sums is not Cauchy and cannot converge.

Answer 2

I can prove its subsequences's convergence but can't find their exact values. $(L-L^{\frac{1}{2}})+(L^{\frac{1}{3}}-L^{\frac{1}{4}})+...+(L^{\frac{1}{n-1}}-L^{\frac{1}{n}}) = L+(-L^{\frac{1}{2}}+L^{\frac{1}{3}})+(-L^{\frac{1}{4}}+L^{\frac{1}{5}})+...-L^{\frac{1}{n}} ＜L-1$

$L+(-L^{\frac{1}{2}}+L^{\frac{1}{3}})+(-L^{\frac{1}{4}}+L^{\frac{1}{5}})+...+(-L^{\frac{1}{n-1}}+L^{\frac{1}{n}}) =(L-L^{\frac{1}{2}})+(L^{\frac{1}{3}}-L^{\frac{1}{4}})+...+L^{\frac{1}{n}} ＞1$