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I think I have the correct answer to the problem specifically, but I'm a little unsure of my usage of the little-o notation, or if this is even what Apostol was suggesting to do. (Apostol does cover little-o with regard to Taylor series, he hasn't mentioned big-O yet). The question in the book (Apostol Calculus Vol. I, Section 10.16 #18) is

Use Gauss' test to prove that the series $$\sum_{n=1}^\infty \left(\frac{1 \cdot 3 \cdot 5\cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} \right)^k$$ converges if $k>2$, and diverges if $k\le2$.

Gauss' test is given in this form: If there is an $N \ge 1$, an $s>1$, and an $M\ge0$ such that

$$\frac{a_{n+1}}{a_n}=1-\frac A n + \frac {f(n)}{n^s}\qquad \forall n\ge N$$ where $|f(n)|\le M$ for all n, then $\sum a_n$ converges if $A>1$ and diverges if $A\le 1$.


My answer: $$\frac {a_{n+1}}{a_n}=\left(\frac{2n+1}{2n+2}\right)^k=\left(1-\frac 1 {2(n+1)}\right)^k$$ Let $x=\frac 1 {n+1}$, then $$\frac{a_{n+1}}{a_n}=\left(1-\frac 1 2 x\right)^k=1-\frac k 2x+o(x)\quad \text{as }x\to0^+$$ I think that this is almost the answer, since it shows that

$$\frac{a_{n+1}}{a_n}=1-\frac k {2n}+o\left(\frac 1 n\right)\qquad \text{ as }n\to +\infty$$ and so I see the correlation with Gauss' formula, and the distinction for $k\le2$ and $k>2$ for divergence and convergence. For some $N$, we have that the function on the right is, at most, some $\epsilon>0$ less than $\frac{a_{n+1}}{a_n}$ but I'm unsure of how to formally proceed.

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1 Answer 1

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Your expansion with $o(1/n^2)$ is wrong. In fact, $$\left(1-\frac 1 2 \frac 1{n+1}\right)^k=1-\frac k{2n}+O\left(\frac1{n^2}\right)=1-\frac An+\frac{f(n)}{n^s}, $$ with $f$ bounded, $A=\dfrac k2$ and $s=2$. The conclusion follows.

Edit (Added to answer some comments by the OP.)

Define $u(x)=(1-\frac12x)^k$, then $u$ is (at least) twice differentiable on $|x|<2$, $u(0)=1$ and $u'(0)=-\frac12k$ hence, for every $x$ such that $|x|<2$, $u(x)=1-\frac12kx+\frac12u''(y)x^2$ for some $y$ such that $|y|\leqslant|x|$. Since $u''$ is bounded on $(0,1)$, say $|u''(z)|\leqslant M$ for every $|z|\leqslant1$, Taylor's formula tells you that $(1-\frac12x)^k=1-\frac12kx+f(x)x^2$ with $|f(x)|\leqslant\frac12M$ for every $|x|\leqslant1$.

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Ah, yes I checked the definition of the Taylor Series expressed in $o$ notation, and my expansion above is definitely incorrect. I edited it accordingly. Apostol has not covered big-O, so I don't know how to use the expansion you provided (or how to derive it). Is there any way to do this with just little-o, or perhaps without any $o$ functions at all? –  process91 Oct 10 '11 at 20:47
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For every $(\varphi(n))$, one may replace $O(\varphi(n))$ by $f(n)\varphi(n)$ with $f$ bounded, the two are equivalent. // Your insistence on using little-o is odd since Gauss's test is expressed in typical big-O form. –  Did Oct 10 '11 at 20:52
    
I don't really personally insist on little-o, it's just that Apostol has not introduced big-O at all. If this problem requires big-O in order to be solved, it must be one of the few I've found which require additional external study (there have only been two or three so far, and I am self-studying here so I don't have anyone else to ask). Does the question require big-O in order to be answered? –  process91 Oct 10 '11 at 20:56
    
As I said: you can replace big-O by the formulation with a bounded $f$, the two are EQUIVALENT. –  Did Oct 10 '11 at 20:59
    
But I have no way to get from $(1-\frac 1 2 \frac 1 {n+1})^k$ to $1-\frac k {2n} + \frac {f(n)} {n^2}$ without using big-O as you described above. –  process91 Oct 10 '11 at 21:01

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