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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!! :-)

CSch of x
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2 Answers2

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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.

Brian M. Scott
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  • Wow nice answer sir! But something is weird, because in my program, it seems to get fixed on a specific number (according to L of course) – CSch of x Aug 03 '20 at 03:07
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    @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 $1.7$. – Brian M. Scott Aug 03 '20 at 03:11
  • 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! – CSch of x Aug 03 '20 at 03:16
  • @StackOMeow: You’re welcome! I don’t know for sure about the odd and even subsequences, but it wouldn’t surprise me. – Brian M. Scott Aug 03 '20 at 06:53
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I can prove its subsequences's convergence but can't find their exact values. $(L-L^{\frac{1}{2}})+(L^{\frac{1}{3}}-L^{\frac{1}{4}})+...+(L^{\frac{1}{n-1}}-L^{\frac{1}{n}}) = L+(-L^{\frac{1}{2}}+L^{\frac{1}{3}})+(-L^{\frac{1}{4}}+L^{\frac{1}{5}})+...-L^{\frac{1}{n}} ＜L-1$

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

JFR
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