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Could I prove that the ratio test still works using $\limsup(\frac{a_{n+1}}{a_n})$ instead of $\lim(\frac{a_{n+1}}{a_n})$? I think for $\limsup<1$ I could show that for $\epsilon>0, N>1 \limsup(\frac{a_{n+1}}{a_n})<1-\epsilon.$ From there I can solve that $\lvert a_k\rvert<(1-\epsilon)^{k-N}\lvert a_N\rvert$ for $k>N$. Thus, by comparison test the left-hand side will converge absolutely. Is this enough to show I can use the ratio test with $\limsup$?

sebastianross
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2 Answers2

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You have the right ideas, but your proof could use just a little tweaking. Let $L = \limsup |a_{n+1}/a_n|$. If $L < 1$, then for $\epsilon := (1 - L)/2$, there exists $N\in \Bbb N$ such that $|a_{n+1}/a_n| < L + \epsilon$ for all $n\ge N$, i.e., $|a_{n+1}| < |a_n|(1+L)/2$ for all $n \ge N$. Thus $|a_n| \le [(1 + L)/2]^{n-N}|a_N|$ for all $n\ge N$. Since $L < 1$, $(1 + L)/2 < 1$, thus $\sum_{n = 1}^\infty [(1 + L)/2]^{n-N}$ converges. Hence, by the comparison test, $\sum_{n = 1}^\infty a_n$ converges absolutely.

kobe
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  • And can this apply as well for L > 1 (being divergent, of course)? I assume I don't need to incorporate liminf, since that's not part of what I'm trying to prove. – sebastianross Feb 25 '15 at 01:07
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    @sebastianross no, you can say that if $\liminf |a_{n+1}/a_n| > 1$, then the series $\sum_{n = 1}^\infty a_n$ diverges. – kobe Feb 25 '15 at 01:10
  • @kobe Maybe its late but why $\sum_{n = 1}^\infty [(1 + L)/2]^{n-N}$ conveges? can you explain me?, maybe because $[(1 + L)/2]^{n-N}<1$ and then apply comparison test? – Wrloord Sep 01 '23 at 01:04
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    @Wrloord the sum $\sum_{n = 1}^\infty [(1 + L)/2]^{n-N}$ is a geometric series, which converges since $|(1 + L)/2| < 1$. – kobe Sep 02 '23 at 01:31
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Yes. You can prove the next statements $1\implies2\implies3$ to obtain the ratio test.

Theorem 1 (Root test). Let $\sum_{n=m}^\infty a_n$ be a series of real numbers, and let $\alpha:=\limsup_{n\to\infty}|a|^{1/n}$.

  1. If $\alpha<1$, then the series $\sum_{n=m}^\infty a_n$ is absolutely convergent.
  2. If $\alpha>1$, then the series $\sum_{n=m}^\infty a_n$ is divergent (not conditional convergent).
  3. If $\alpha=1$, we cannot assert any conclusion.

Lemma 2. Let $(a_n)_{n=m}^\infty$ be a sequence os positive numbers. Then we have $$\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}\le\liminf_{n\to\infty}a_n^{1/n}\le\limsup_{n\to\infty}a_n^{1/n}\le\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}.$$

Corollary 3 (Ratio test). Let $\sum_{n=m}^\infty a_n$ be a series of non-zero numbers. (The non-zero hypothesis is required so that the ratios $|a_{n+1}|/|a_n|$ appearing below are well-defined.)

  1. If $\limsup_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}<1$, then the series $\sum_{n=m}^\infty a_n$ is absolutely convergent.
  2. If $\liminf_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}>1$, then the series $\sum_{n=m}^\infty a_n$ is divergent (not conditional convergent).
  3. In the remaining cases, we cannot assert any conclusion.
Cristhian Gz
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