Can we still use the ratio test if the sequence of ratios diverges to infinity?

[kingwinner]

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

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

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

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


Thanks for clarifying!

 

[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&gt;M for all k>N. That means that for k>N, we have |a_k|&gt;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}&gt;M, i.e. |a_k|&gt;M^k. Again, clearly divergent.

 

[kingwinner]

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

c=∞>1 => ∑ak diverges?

 

[Tinyboss]

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

"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.