AMM 12181-05 issue

The original proposed issue was to prove $\sum_{k=2}^{\infty}\frac{1}{k}\int_0^{\infty}\left\{\frac{1}{\sqrt[k]{x}}\right\}dx=\gamma$

Here are two proposed solutions of mine

Solution 1 . Call the fractional integral $\displaystyle J_k= \int_0^1\left\{\frac{1}{\sqrt[k]{x}}\right\}dx$ for all $k\geq 2$ . We shall prove that the integral $\displaystyle J_k = \frac{k}{k-1}-\zeta(k)$ . So we substitute $x=u^{-k}$ then we have the integral above as $k\int_1^{\infty}\frac{\left\{u\right\}}{u^{k+1}}du=k\sum_{n\geq 1 }\int_n^{n+1}\frac{u-n}{u^{k+1}}du=k\sum_{n\geq 1}\int_{n}^{n+1}\left(\frac{1}{u}-\frac{n}{u^{k+1}}\right)du$ Evaluating the integrals and simplification yields $J_k=-\frac{k}{k-1}\underbrace {\sum_{n\geq 1}\left(\frac{1}{(n+1)^{k-1}}--\frac{1}{n^{k-1}}\right)}_{\textbf{-1}}+ {\sum_{n\geq 1}\left(\frac{1}{(n+1)^{k}}-\frac{1}{n^{k}}\right)n}$ Further note that the former sum (underbraced) is telescoping series that converges to $-1$ this we have former sum is equal to $\frac{k}{k-1}\cdots (1)$ and latter sum we write as $\sum_{n\geq 1}\left(\frac{1}{(n+1)^{k-1}}-\frac{1}{n^{k-1}}-\frac{1}{(n+1)^{k}}\right)=\sum_{n\geq 1}\left(-1-\frac{1}{(n+1)^{k}}\right)=-\zeta(k)\cdots(2)$ Thus from $(1)$ and $(2)$ we have $\displaystyle J_k=\frac{k}{k-1}-\zeta(k)$ . Now we are left to evaluate $\sum_{k\geq 2}J_k= \sum_{k\geq 2}\frac{1}{k}\left(\frac{k}{k-1}-\zeta(k)\right)=\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k}-\left(\frac{\zeta(k)-1}{k}\right)\right)$ here the sum $\displaystyle \sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k}\right)=\lim_{m\to\infty}\sum_{k=1}^m\left(\frac{1}{k-1}-\frac{1}{k}\right)=\lim_{m\to\infty}\left(1-\frac{1}{m+1}\right)=1$ and to evaluate the sum $\displaystyle \sum_{k\geq 2}\frac{\zeta(k)-1}{k}\cdots(3)$ recall that $\displaystyle \zeta(k)-1=\sum_{m=2}^{\infty}\frac{1}{m^k}$ . plugging in $(3)$ gives us $\sum_{k=2}^{\infty}\frac{\zeta(k)-1}{k}=\sum_{k=2}^{\infty}\frac{1}{k}\left(\sum_{m=2}^{\infty}\frac{1}{m^k}\right)=\sum_{m=2}^{\infty}\left(\sum_{k\geq 2}\frac{1}{k m^k}\right)=\sum_{m\geq 2}\left(-\frac{1}{m}-\ln\left(\frac{m-1}{m}\right)\right)$ the latter sum further can be written as $-\sum_{m\geq 2}\left(\frac{1}{m}-\ln m+\ln(m-1)\right)=-\lim_{m\to\infty}\sum_{j= 2}^m\left(\frac{1}{j}+\ln(j-1)-\ln j\right)$ Note that $\displaystyle \sum_{j=2}^{m} \frac{1}{j}= H_m-1$ and $\displaystyle \sum_{j=2}^m (\ln(j-1)-\ln j)= 0-\ln m=-\ln m$ therefore we yield $-\lim_{m\to\infty}\sum_{j=2}^m\left(\frac{1}{j}+\ln(j-1)-\ln j\right)=-\lim_{m\to\infty} (H_m-\ln m-1)=1-\gamma\cdots(4)$ subtracting $3$ from $4$ we have the desired result $\displaystyle \sum_{k\geq 2}J_k=\gamma$ .

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Solution 2 Let $f(x)=\left\{\frac{1}{\sqrt[k]{x}}\right\}$ for all $x\in (0,1]$ and $k\geq 2$

$\int_0^1f(x)dx = \int_0^1\left(\frac{1}{\sqrt[k]{x}}-\left\lfloor \frac{1}{\sqrt[k]{x}}\right\rfloor\right)dx=\frac{k}{k-1}-I(k)$ We note that $n\leq x < n+1$ and hence $\sqrt[k]{n} \leq \sqrt[k]{x} < \sqrt[k]{n+1}$ which implies $\frac{1}{\sqrt[k]{n+1}} < \frac{1}{\sqrt[k]{x}}\leq \frac{1}{\sqrt[k]{n}}$ . also on the interval $( (n+1)^{-1/k}, n^{-1/k}], \lfloor x^{-1/k}\rfloor = n$ . Therefore $I_k = -n\sum_{n\geq 1}\int_{\frac{1}{n^k}}^{\frac{1}{(n+1)^k}}dx=-\sum_{n\geq 1} \left(\frac{1}{(n+1)^k}-\frac{1}{n^k}\right)n=\zeta(k)$ giving us $\int_0^1 f(x)dx =\frac{k}{k-1} -\zeta(k)$ To prove $\displaystyle \sum_{k\geq 2} \int_0^1 f(x) dx = \sum_{k\geq2} \left(\frac{1}{k-1}-\frac{1}{k}-\frac{\zeta(k)-1}{k}\right)=\gamma$ As we have in solution 1 $\displaystyle \sum_{k\geq 2} \left(\frac{1}{k-1}-\frac{1}{k}\right)=1\cdots (5)$ and to show that $\displaystyle\sum_{k\geq}\frac{\zeta(k)-1}{k}=1-\gamma$ we make an alternative approach as follows. We have that $\frac{\zeta(k)-1}{k}=\int_0^{\infty}\frac{x^k}{k !}\cdot \frac{dx}{xe^x(e^x-1)}dx$ and hence on summing we have $\sum_{k\geq 2}\frac{\zeta(k)-1}{k}=\int_0^{\infty}\frac{e^x-x-1}{xe^x(e^x-1)}dx=\int_0^{\infty}\frac{dx}{e^x}-\int_{0}^{\infty}\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)dx=1-A$ since $A=\displaystyle -\int_{0}^{\infty}\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)dx=-\gamma$ and and hence $\displaystyle\sum_{k\geq 2}\frac{\zeta(k)-1}{k}=1-\gamma\cdots(6)$ and hence from $(5)$ and $(6)$ we proved $\displaystyle \sum_{k\geq 2}\frac{1}{k}\int_0^1 f(x)dx=\gamma$ .