Let us determine whether the following series converge, converge absolutely, converge conditionally, or diverge.
1) "\\displaystyle\\sum_{k=1}^{\\infty}\\frac{3^k}{k^22^k}"
Let us use D'Alembert's Ratio Test.
Since "\\lim\\limits_{n\\to\\infty}(\\frac{3^{k+1}}{(k+1)^22^{k+1}}:\\frac{3^k}{k^22^k})\n=\\lim\\limits_{n\\to\\infty}\\frac{3k^2}{2(k+1)^2}=\\lim\\limits_{n\\to\\infty}\\frac{3k^2}{2(k^2+2k+1)}=\\frac{3}2>1,"
we conclude that this series diverge.
2) Since
"\\displaystyle\\sum_{k=1}^{\\infty}\\frac{(-1)^{k+1}}{ \\sqrt{k+1} -\\sqrt{k}}\n=\\displaystyle\\sum_{k=1}^{\\infty}\\frac{(-1)^{k+1}(\\sqrt{k+1} +\\sqrt{k})}{ (\\sqrt{k+1} -\\sqrt{k})(\\sqrt{k+1} -\\sqrt{k})}\n=\\displaystyle\\sum_{k=1}^{\\infty}\\frac{(-1)^{k+1}(\\sqrt{k+1} +\\sqrt{k})}{ ((k+1)-k)}\n=\\displaystyle\\sum_{k=1}^{\\infty}(-1)^{k+1}(\\sqrt{k+1} +\\sqrt{k}),"
and "\\lim\\limits_{k\\to\\infty}(\\sqrt{k+1} +\\sqrt{k})=\\infty,"
we conclude that the necessary condition of covergence is not met, and hence this series is divergent.
3) "\\displaystyle\\sum_{k=1}^{\\infty}(-1)^{k+1}( \\sqrt{k+1} -\\sqrt{k})"
Consider the series "\\displaystyle\\sum_{k=1}^{\\infty}|(-1)^{k+1}( \\sqrt{k+1} -\\sqrt{k})|=\\displaystyle\\sum_{k=1}^{\\infty}(\\sqrt{k+1} -\\sqrt{k})."
Since "A_n=(\\sqrt{2} -1)+(\\sqrt{3} -\\sqrt{2})+(\\sqrt{4} -\\sqrt{3})+\\ldots +(\\sqrt{n+1} -\\sqrt{n})=\\sqrt{n+1} -1,"
and "\\lim\\limits_{n\\to\\infty}A_n=\\lim\\limits_{n\\to\\infty}(\\sqrt{n+1} -1)=\\infty," we conclude that the series "\\displaystyle\\sum_{k=1}^{\\infty}(-1)^{k+1}( \\sqrt{k+1} -\\sqrt{k})" diverge absolutely.
Comments
Leave a comment