# Find the domain in which a sequence converges

I seem to have a lot of problems in the format of:

Given the sequence $a_n=\frac{n^\alpha+1}{2n^8+5}$, for $\alpha > 0$, find the values of $\alpha$ such that the sequence converges.

However, I can never seem to get Mathematica to solve for a variable in a limit.

I have tried

Solve[Limit[(1 + n^a)/(5 + 2 n^8), n -> Infinity, Assumptions :> a > 0] == 0, a]

Reduce[Limit[(1 + n^a)/(5 + 2 n^8), n -> Infinity, Assumptions :> a > 0] == 0, a]


Solve gave me no solutions, with a warning about inverse function. Reduce gave me the solution is C[1] ∈ Integers && α == 0, which also violates the initial assumption ($\alpha > 0$). However, I tried particular cases manually, such as

Limit[(1 + n^3)/(5 + 2 n^8), n -> Infinity, Assumptions :> a > 0] == 0


which gave True. Am I going about this in the wrong way?

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Hey! The denominator has a n^8 in it ... Doesn't that whisper something near your left ear? – belisarius has settled Apr 10 '13 at 2:36
@belisarius Yes, I'm aware of the solution to the problem, I was just wondering if it is possible to have mathematica evaluate these types of problems involving limits. – Paul Spinelli Apr 10 '13 at 2:42
Expanding the function around $n=\infty$ with Series more or less displays the solution. – whuber Apr 10 '13 at 3:04
@whuber Hmm do you mean you just used: Series[a[n], {n, Infinity,Infinity}? – Paul Spinelli Apr 10 '13 at 3:23
I don't think so, Paul, because MMA will not evaluate the limit symbolically. – whuber Apr 10 '13 at 15:17

For some reasons Limit does not work well with inequalities in assumptions. To see the problem one can try this :

Manipulate[ Limit[(1 + n^a)/(5 + 2 n^8), n -> Infinity, Assumptions -> a == h],
{h, 0, 20}]


it does work well unlike :

Manipulate[ Limit[(1 + n^a)/(5 + 2 n^8), n -> Infinity, Assumptions -> a > h],
{h, 0, 20}]


Since the first way works well, one can find easily (also in more sophisticated examples) appropriate values, e.g.

Limit[(1 + n^a)/(5 + 2 n^8), n -> Infinity, Assumptions -> #] & /@ { a < 8, a == 8, a > 8}

 {0, 1/2, Infinity}


The problem with using Limit with Reduce is rather of more general nature, e.g. :

Reduce[ Limit[(1 + n^a)/(5 + 2 n^8), n -> Infinity, Assumptions -> a == h] == 0, h, Reals]

 False


This is certainly a bug.

On the other hand SumConvergence seems to behave better :

Assuming[a ∈ Reals, SumConvergence[(1 + n^a)/(5 + 2 n^8), n]]

  a < 7

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