# Finding a closed formula for a sum

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!! :-)

• Does it converge? Aug 3 '20 at 3:01
• @JackyChong Well in my program I checked values $k = 1,000,000+$ and it seems to converge but I am not sure.. Aug 3 '20 at 3:02
• Have you tried the divergence test? Aug 3 '20 at 3:02
• Individual terms converge to $1$ not $0$. Aug 3 '20 at 3:05
• I found that If this sum converged, your partial sums too.$\sum_{k=1}^{\infty} L^{\frac{1}{k}} -1$ Aug 3 '20 at 3:31

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.
• @StackOMeow: I would check the program; for $L=2$ the partial sums for a modest number of terms should appear to oscillate between values not too far from $0.7$ and $1.7$. Aug 3 '20 at 3:11
• Ohh right! it is indeed, for some even values of $k$ the sum = $c$ and for odd $k$ equals $c+1$ . So will we be able to find a closed form of the partial sums for even / odd $k$'s ? thank you sir! Aug 3 '20 at 3:16
I can prove its subsequences's convergence but can't find their exact values. $$(L-L^{\frac{1}{2}})+(L^{\frac{1}{3}}-L^{\frac{1}{4}})+...+(L^{\frac{1}{n-1}}-L^{\frac{1}{n}}) = L+(-L^{\frac{1}{2}}+L^{\frac{1}{3}})+(-L^{\frac{1}{4}}+L^{\frac{1}{5}})+...-L^{\frac{1}{n}} ＜L-1$$
$$L+(-L^{\frac{1}{2}}+L^{\frac{1}{3}})+(-L^{\frac{1}{4}}+L^{\frac{1}{5}})+...+(-L^{\frac{1}{n-1}}+L^{\frac{1}{n}}) =(L-L^{\frac{1}{2}})+(L^{\frac{1}{3}}-L^{\frac{1}{4}})+...+L^{\frac{1}{n}} ＞1$$