proof of ratio test

Assume $k<1$ . By definition $\\exists N$ such that
$n>N\\implies|\\frac{a_{n+1}}{a_{n}}-k|<\\frac{1-k}{2}\\implies|\\frac{a_{n+1}}{a_{n%}}|<\\frac{1+k}{2}<1$

i.e. eventually the series $|a_{n}|$ becomes less than a convergent geometric series, therefore a shifted subsequence of $|a_{n}|$ converges by the comparison test. Note that a general sequence $b_{n}$ converges iff a shifted subsequence of $b_{n}$ converges. Therefore, by the absolute convergence theorem, the series $a_{n}$ converges.

Similarly for $k>1$ a shifted subsequence of $|a_{n}|$ becomes greater than a geometric series tending to $\\infty$ , and so also tends to $\\infty$ . Therefore $a_{n}$ diverges.

单词	ProofOfRatioTest
释义	proof of ratio test Assume $k<1$ . By definition $\\exists N$ such that $n>N\\implies\|\\frac{a_{n+1}}{a_{n}}-k\|<\\frac{1-k}{2}\\implies\|\\frac{a_{n+1}}{a_{n%}}\|<\\frac{1+k}{2}<1$ i.e. eventually the series $\|a_{n}\|$ becomes less than a convergent geometric series, therefore a shifted subsequence of $\|a_{n}\|$ converges by the comparison test. Note that a general sequence $b_{n}$ converges iff a shifted subsequence of $b_{n}$ converges. Therefore, by the absolute convergence theorem, the series $a_{n}$ converges. Similarly for $k>1$ a shifted subsequence of $\|a_{n}\|$ becomes greater than a geometric series tending to $\\infty$ , and so also tends to $\\infty$ . Therefore $a_{n}$ diverges.
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