[tex]a_n=\frac{\sqrt[n+k+1]{n+1}-\sqrt[n+k]{n}}{\sqrt[n+k]{n+1}-\sqrt[n+k]{n}}=\frac{e^{-\frac{\ln n}{(n+k)(n+k+1)}}e^{\frac{\ln(1+\frac1n)}{n+k+1}}-1}{e^{\frac{\ln(1+\frac1n)}{n+k}}-1}=\\=\frac{\bigg(\Big(1-\frac{\ln n}{(n+k)(n+k+1)}+O(\frac{\ln^2n}{n^4})\Big)\Big(1+\frac1{n(n+k+1)}+O(\frac1{n^3})\Big)-1\bigg)}{1+\frac1{n(n+k)}+O(\frac1{n^3})-1}=1-\frac{n\ln n}{n+k}+O\left(\frac{\ln n}n\right)\\[/tex][tex]S_n=\sum\limits_{k=1}^na_n=n\,-\,n\ln n\ln 2\,+\,O(\ln n)\wedge \frac{S_n}{n}=1\,-\,\ln 2\ln n\,+\,O\left(\frac{\ln n}n\right)\,\to -\infty[/tex]
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[tex]a_n=\frac{\sqrt[n+k+1]{n+1}-\sqrt[n+k]{n}}{\sqrt[n+k]{n+1}-\sqrt[n+k]{n}}=\frac{e^{-\frac{\ln n}{(n+k)(n+k+1)}}e^{\frac{\ln(1+\frac1n)}{n+k+1}}-1}{e^{\frac{\ln(1+\frac1n)}{n+k}}-1}=\\=\frac{\bigg(\Big(1-\frac{\ln n}{(n+k)(n+k+1)}+O(\frac{\ln^2n}{n^4})\Big)\Big(1+\frac1{n(n+k+1)}+O(\frac1{n^3})\Big)-1\bigg)}{1+\frac1{n(n+k)}+O(\frac1{n^3})-1}=1-\frac{n\ln n}{n+k}+O\left(\frac{\ln n}n\right)\\[/tex][tex]S_n=\sum\limits_{k=1}^na_n=n\,-\,n\ln n\ln 2\,+\,O(\ln n)\wedge \frac{S_n}{n}=1\,-\,\ln 2\ln n\,+\,O\left(\frac{\ln n}n\right)\,\to -\infty[/tex]