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As a part of a problem I am using this:

I know that $\sum_{k=n+1}^\infty a_kz^k$ converges absolutely in a region around zero, offcourse its value at zero is zero. I also know that it is is continuous.

But I need to show that as z goes to zero I have:

$$\left|\sum_{k=n+1}^\infty a_kz^k\right|\le C\cdot|z^{n+1}|$$

I mean on the left side I have:

$|\sum_{k=n+1}^\infty a_kz^k|=|a_{n+1}\cdot z^{n+1}+\sum_{k=n+2}^\infty a_kz^k|$. I really feel that there is some simple or smart trick to just show that we can pick a C that covers the last part of the sum. But I have no idea of how to find it. Can someone please think of a trick here?

Michael Hardy
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user119615
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    If you denote the function by $f$, what is $$g(z) = \frac{f(z)}{z^{n+1}},?$$ – Daniel Fischer Sep 12 '14 at 20:44
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    Please don't use the asterisk $*$ for multiplication (outside of programming). In mathematics, it stands for convolution, which is a different sort of beast. I replaced it with a dot (\cdot). – Harald Hanche-Olsen Sep 12 '14 at 20:48
  • @DanielFischer I get $\Sigma_{i=n+1}^{\infty}a_k*z^{i-n-1}$. I see what you mean. This is probably a very stupid question, but how do we know it converges to zero? It probably is obvious, but I can't use l'hopital. Is this argument correct?: g is continuos at all points around zero, because it is created by two continuous functions. If we look at the expression for g it is zero at zero. [But I can't really finish because I dont see how to force g to converge to zero.] – user119615 Sep 12 '14 at 20:54
  • @HaraldHanche-Olsen Thank you, I will try to remember that! – user119615 Sep 12 '14 at 20:54
  • Is this the tail of a power series of an analytic function? – Mhenni Benghorbal Sep 12 '14 at 20:56
  • The series for $g$ converges in the same disk as the series for $f$ [the convergence in boundary points of the disk may differ, but we're not interested in that]. The sum of a convergent power series is a continuous function on its disk of convergence. Thus we have $g(0) = a_{n+1}$, and $\lvert g(z)\rvert \leqslant \lvert a_{n+1}\rvert + 1$ in a neighbourhood of $0$. – Daniel Fischer Sep 12 '14 at 20:57
  • @MhenniBenghorbal Yeah, but how do we know it is analytic after the division? I mean, we divided on something and that may mess things up because what we divided on is not defined at zero? I mean after we look at the expression we see that at zero it is zero, but I don't see how convergence to zero is guranteed? – user119615 Sep 12 '14 at 20:58
  • @DanielFischer Do we use that in the new series $1/R=lim sup |a_n|^{1/n}$ doesnt change? I mean technically it could change because the new $a_1$ doesnt correspond to the original n=1. But we can prove it is the same either way? – user119615 Sep 12 '14 at 21:01
  • What you seem to be missing is that the exponent $i-n-1$ is nonnegative in all terms of the new series. No nasty denominators to worry about, it's just a power series. – Harald Hanche-Olsen Sep 12 '14 at 21:03
  • Basically what I mean is I have to prove that $lim sup |a_{i+n+1}|^{1/i}=lim sup|a_i|^{1/i}$? – user119615 Sep 12 '14 at 21:06
  • That's one way to see that the two series have the same radius of convergence. Another way is to use that fact that the power series $\sum_{k=0}^\infty b_k z^k$ has radius of convergence $\geqslant r$ if and only if $a_k\cdot r^k$ is a bounded sequence. – Daniel Fischer Sep 12 '14 at 21:06
  • @user119615: See my answer. – Mhenni Benghorbal Sep 12 '14 at 21:09
  • @HaraldHanche-Olsen I understand that, the problem is that we have a completely new powerseries, so we do not even know if it converges I think, I mean it is not just the tail of a power series that we know converges, we have change it by mixing the correspening indexes in the $a_n$ and exponent? But I see it will follow from the lim sup argument, or with what Daniel Fisher said about r(I dont see why that is equivalent with the lim sup argument, so I would have to prove that first). So we have to do the limsup argument?, it seems very difficult. – user119615 Sep 12 '14 at 21:19

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Basically you want to prove the following

$$ \sum_{k=n+1}^{\infty}a_kz^k = O(z^{n+1}), \quad z\to 0. $$

Here is a useful result you can use

If $\lim_{z\to z_0}\frac{f(z)}{g(z)} = b $ then $f(z)=O(g(z)) $ or $|f(z)|\leq M |g(z)|$,

where $b$ is finite. See the main result.

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