# Root test & Ratio test

1. Mar 8, 2010

### kingwinner

Let ∑ak be a series with positive terms.
Ratio test:
Suppose ak+1/ak -> c.
If c<1, then ∑ak converges.
If c>1, then ∑ak diverges.
If c=1, the test is inconclusive.

What if ak+1/ak diverges (i.e. ak+1/ak->∞)? Do we count this as falling into the case c>1? Can we say whether ∑ak converges or not?

Root test:
Suppose limsup (ak)1/k = r.
If r<1, then ∑ak converges.
If r>1, then ∑ak diverges.
If r=1, the test is inconclusive.

What if limsup (ak)1/k = ∞? Do we count this as falling into the case r>1? Can we say whether ∑ak converges or not?

Thanks for clarifying!

2. Mar 8, 2010

### Tinyboss

Just unwind the definitions, and it should be clear.

First question: if $$a_{k+1}/a_k\to\infty$$, then for M as big as you like, there's some N such that $$a_{k+1}/a_k>M$$ for all k>N. That means that for k>N, we have $$|a_k|>M^{k-N}$$. Clearly that's diverging.

Second question: if $$\text{lim sup }a_k^{1/n}=\infty$$, then for M and N as large as you'd like, there's a k>N with $$a_k^{1/k}>M$$, i.e. $$|a_k|>M^k$$. Again, clearly divergent.

3. Mar 9, 2010

### kingwinner

So in the ratio test, the case c=∞ is allowed?

c=∞>1 => ∑ak diverges?

4. Mar 9, 2010

### Tinyboss

"The" ratio test requires that the sequence of ratios converges--it's right there in the definition you gave. If the sequence diverges to infinity, though, a simple modification of the argument yields the same result.

Whether that distinction is meaningful is up to you.