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Let $\sum c_n$ be a series of positive numbers. Assume that $$\lim_n {{c_{n+1}}\over {c_n}} = r$$

If $0\lt r\lt 1$ then the series converges ; if $r\gt 1$ then diverges .

Now , if $r\gt 1$ then we have $$c_n \ge c_N$$ for $\forall n\ge N$ .

How do we have this $?$ Because , using the definition of limit , I got $$c_{n+1}\lt ({r+\epsilon})c_n$$.

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  • $\begingroup$ 1. What's your question? 2. The definition of limit tells you that for any $\varepsilon$, there exists an $N$ such that for any $n \geq N, (r-\varepsilon) c_n < c_{n+1} < (r+\varepsilon) c_n$. It is the left inequality and not the right that gives you what you need. $\endgroup$ – Brian Tung Nov 17 '15 at 18:19
  • $\begingroup$ @BrianTung : Yes , that is my problem. Why do we have to consider the left one this time while for the $r\lt 1$ part we considered the right inequality $?$ Please explain that reason. $\endgroup$ – user118494 Nov 17 '15 at 18:22
  • $\begingroup$ Because when $r > 1$, then what you want to show is that beyond a certain index $N$, everything gets bigger. We show this by setting $\epsilon < r-1$; then we define $r' \equiv r-\varepsilon > 1$ and the left inequality tells you that $c_{n+1} > r' c_n$. In other words, the $\{c_n\}$ diverge geometrically beyond $N$. When $r < 1$, we want to show that everything gets smaller, so we use the right inequality, and define $r'' \equiv r+\varepsilon < 1$, and then $c_{n+1} < r'' c_n$; the $\{c_n\}$ converge geometrically beyond $N$. $\endgroup$ – Brian Tung Nov 17 '15 at 18:27
  • $\begingroup$ For example, suppose $r = 2$. Then the definition of the limit says that if we choose (say) $\varepsilon = 1/2$, then $3/2 < c_{n+1}/c_n < 5/2$ for $n \geq N$ (for some $N$). We focus on the left inequality, which is equivalent to $c_{n+1} > (3/2) c_n$, so the $\{c_n\}$ diverge geometrically at a rate faster than $3/2$. $\endgroup$ – Brian Tung Nov 17 '15 at 18:29
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    $\begingroup$ As $r + \epsilon$ can be larger than 1, than $c_{n+1} < (\text{something larger than 1)}\cdot c_n$. Doesn't tell us anything of any use. $\endgroup$ – fleablood Nov 17 '15 at 18:29
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Because $r > 1$ there is $N \in \mathbb N$ so that $\frac{c_{n+1}}{c_{n}} \gt 1 , \forall n \ge N$. But $$\frac{c_{n}}{c_{N}} = \frac{c_{n}}{c_{n-1}}\frac{c_{n-1}}{c_{n-2}}...\frac{c_{N+1}}{c_{N}} \gt 1...1 = 1$$ for all $n>N$

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"$c_n \ge c_N$ for $\forall n\ge N$" and "$c_{n+1}\lt ({r+\epsilon})c_n$" are not at all contradictory or in conflict. As $r+ \epsilon > r > 1$, we can always have $c_n \lt c_{n+1} \lt (r + \epsilon)c_n$. Indeed $M_i\cdot c_N > c_{N + i} \approx r^i C_N > C_N$ for some $M_i > 1$.

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