Integral test vs root test vs ratio test

I want to know the difference between ratio test and root test.

That is, I want to know the series which is tested by ratio but not root.

Or tested by root but not root.

As you know neither the ratio nor the root test help with $p$-series, i.e., $\sum_{n=1}^\infty \frac{1}{n^p}$ $( p > 1 )$.

Nevertheless we have a difficulty in testing $\sum_{n=1}^\infty n^2 e^{-n}$ by integral test. Surely we know by experience that the series converges.

But to prove we may use ratio or root test.

For this convenience, I think that root and ratio test is powerful.

But I do not know the difference between them in testing : As far as I know, the convergent series that is tested by root can be tested by ratio and vice versa.

Question : Is there a convergent series that is tested by root and cannot be tested by ratio ? Or is there a convergent series that is tested by ratio and cannot be tested by root ?

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Have a look at this. –  Ｊ. Ｍ. Aug 6 '12 at 13:20
@J.M. That's a nice article! What a very clever way to see directly the relationship between the two series tests. –  Morgan Sherman Aug 6 '12 at 16:39
@Morgan, neat, right? I had been looking for an opportunity to link to it for a while now... –  Ｊ. Ｍ. Aug 6 '12 at 16:46
The link does not work for me. Could you please post the reference or MathSciNet number? Thanks in advance. –  Shripad Garge Jul 24 '14 at 14:20

Root and ratio test can only recognize a series as convergent if its terms decrease at least exponentially in absolute value. For what it's worth I prefer the following version of this principle:

If there is a $q<1$, a $C$ and a $k_0$ such that $|a_k|\leq C q^k$ for all $k>k_0$ then the series $\sum_{k=0}^\infty a_k$ converges absolutely.

The root test (or the test just quoted) is a little bit stronger than the ratio test, inasmuch the decrease of the $|a_k|$ only has to be exponential "on average" and not at each step. Consider in this regard the series $$\sum_{k=0}^\infty a_k:={1\over2}+{1\over1}+{1\over8}+{1\over4}+{1\over32}+{1\over16}+\ldots\quad.$$ Here the quotients are alternatively $2$ and ${1\over8}$; and the ratio test remains undecisive. In reality one has $|a_k|\leq 2^{1-k}$ which guarantees convergence.

Series whose general term $a_k$ is of order ${1\over k^p}$ for some $p>1$ converge much slower than the above series, and you need the integral test, or a comparison of the form $$|a_k|\leq {C\over k^p}\qquad(k>k_0)$$ to prove that they are convergent.

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I'm not sure that lack of monotonicity is the problem with the ratio test. For example consider the series $1 + 1\cdot\frac23 + \frac12 + \frac12\cdot\frac23 + \frac1{2^2} + \frac1{2^2}\cdot\frac23 + \ldots$. –  Morgan Sherman Aug 6 '12 at 16:24
@Morgan Sherman: Your example passes the ratio test with $q:={3\over4}$; but I have have made my statement more precise nevertheless. –  Christian Blatter Aug 6 '12 at 17:31
I believe $a_{n+1}/a_n$ is alternately 3/4 or 2/3, depending on whether $n$ is even or odd (so the limit fails to exist). Maybe I haven't made the example clear: take $a_{2n} = \frac1{2^n}$ and $a_{2n+1} = \frac1{2^n}\cdot\frac23$. Then my example is $\sum_{n=0}^{\infty} a_n$. –  Morgan Sherman Aug 6 '12 at 17:38
@Morgan Sherman: Yes, but all quotients are $\leq{3\over4}<1$, and this is sufficient. For the ratio test it is not necessary that the quotients have a limit $q<1$. –  Christian Blatter Aug 6 '12 at 17:47
Oh I see, the ratio test I'm familiar with calculates a limit, not a limsup. Thanks for clarifying. –  Morgan Sherman Aug 6 '12 at 17:55