32 . Ratio Test

22 - Ratio Test

Ratio Test

Let $(a_k)$ be a real sequence of positive terms
Assume that $\frac{a_{k+1}}{a_k}\to L$ as $k\to \infty$

1. If $0\le L<1$, then $\sum a_k$ converges
2. If $K>1$ then $\sum a_k$ diverges

If $L=1$ or $L$ doesn’t exist, cannot use test must show manually

Proof

1. Assume that $0\le L<1$
   Let $\alpha =\frac{1+L}{2}$, then $L<\alpha <1$
   Let $ϵ=\alpha -L>0$
   Since $\frac{a_{k+1}}{a_k}\to L$, there is $N$ such that

$$\text{if }k\ge N\text{ then }\left|\frac{a_{k+1}}{a_k}-L\right|<ϵ$$

Hence $\frac{a_{k+1}}{a_k}<L-ϵ=\alpha$
For $k\ge N$ we have $0<a_k\le {\alpha }^{k-N}{a}_N$
However ${\sum }_{k\ge N}{\alpha }^{k-N}{\alpha }_N$ converges (geometric series with $|\alpha |<1$)
Hence by Comparison Test then $\sum a_k$ converges
Remember that the first $N$ terms don’t affect convergence
2) Assume that $L>1$

Case $l\in \mathbb{R}$
Let $\alpha =\frac{1+L}{2}$ so $1<\alpha <L$
Let $ϵ=L-\alpha >0$
Since $\frac{a_{k+1}}{a_k}\to L$, there is $N$ such that

$$\text{if }k\ge N\text{ then }\left|\frac{a_{k+1}}{a_k}-L\right|<ϵ$$

Hence $\frac{a_{k+1}}{a_k}>L-ϵ=\alpha$
For $k\ge N$, then $a_k\ge {\alpha }^{k-N}{a}_N>0$
So $a_k\nrightarrow 0$ as $k\to \infty$, so $\sum a_k$ diverges

Case $L=\infty$
Let $\alpha =2$
As $\frac{a_{k+1}}{a_k}\to \infty$, there is $N$ such that

$$\text{if }k\ge N\text{ then }\frac{a_{k+1}}{a_n}>\alpha$$

Then using Case $L\in \mathbb{R}$ it diverges

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Ratio Test with Absolute Convergence

Let $(a_k)$ be a sequence of non-zero (real of complex) numbers
Assume that $\left|\frac{a_{k+1}}{a_k}\right|\to L\text{ as }k\to \infty$

1. If $0\le K<1$ then $\sum a_k$ converges absolutely and hence converges
2. If $K>1$ then $\sum a_k$ diverges

Similar to the ratio test if $L=1$ then you have to use another method

Proof (Not Detailed)

1. Apply Ratio Test to $(|a_k|)$
2. If $L>1$ then the Ratio Test applied to $(|a_k|)$ shows that

$$|a_j|\nrightarrow 0\text{ so }{a}_k\nrightarrow 0$$

Hence $\sum a_k$ diverges

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