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I have done some manipulation and got that $$\frac{1}{1+e^z} = \sum_{n=0}^\infty \frac{n!}{n!+z^n}$$ by the fact that:

$$\frac{1}{1+e^z}= \frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}=\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{2}}+\ldots = \frac{1}{2}+\frac{1}{1+z}+\frac{2!}{2!+z^2}+\frac{3!}{3!+z^3}+\ldots$$ $$= \sum_{n=0}^\infty\frac{n!}{n!+z^n}$$

Assuming I did the above right, I am having trouble finding the radius of convergence of the Taylor series given above. I tried the ratio test but got stuck.

Edit: I just realized I did this completely wrong, thinking that I was taking $$\sum\frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}$$.

Could anyone help me? My question wants me to compute the first four terms of the taylor series for $\frac{1}{1+e^z}$ and find the radius of convergence. Perhaps they do not want me to actually find the explicit form?

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    $\begingroup$ The series you have written down aren't Taylor series, which must be of the form $\sum_{n = 0}^{\infty} a_n z^n$. $\endgroup$ – Travis Willse Mar 25 '15 at 6:19
  • $\begingroup$ @Travis I understand, that's why I wrote the edit, saying I've done it completely wrong. $\endgroup$ – H5159 Mar 25 '15 at 6:20
  • $\begingroup$ I see what you mean now, I'll give an answer sketching a straightforward method that's a little more pleasant than computing the fourth derivative of the given function. $\endgroup$ – Travis Willse Mar 25 '15 at 6:21
  • $\begingroup$ $e^z=-(-e^z)$. Try writing this is a geometric series. $\endgroup$ – user170231 Mar 25 '15 at 6:25
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I assume you want the Maclaurin series, i.e. the Taylor series about $0$.

Write $$1 + e^z = 2 (1 + Q(z))$$
where $$Q(z) = \dfrac{z}{2\cdot 1!} + \dfrac{z^2}{2\cdot 2!} + \dfrac{z^3}{2\cdot 3!} + \ldots $$ So $$ \dfrac{1}{1+e^z} = \dfrac{1}{2(1+Q(z))} = \dfrac{1}{2} \left( 1 - Q(z) + Q(z)^2 - Q(z)^3 + \ldots \right) $$ For $n \ge 1$, the $z^n$ term in the Taylor series comes from the $Q(z)^j$ terms for $j = 1$ to $n$. Thus the series is $\sum_{n=0}^\infty a_n z^n$ where $$ \eqalign{ a_0 &= \dfrac{1}{2}\cr a_1 &= \dfrac{1}{2} \left( -\dfrac{1}{2\cdot 1!}\right) = -\dfrac{1}{4}\cr a_2 &= \dfrac{1}{2} \left( - \dfrac{1}{2 \cdot 2!} + \dfrac{1}{(2 \cdot 1!)^2} \right) = 0\cr a_3 &= \dfrac{1}{2} \left( -\dfrac{1}{2 \cdot 3!} + \dfrac{2}{(2\cdot 1!)(2\cdot 2!)} - \dfrac{1}{(2 \cdot 1!)^3}\right) = \dfrac{1}{48}\cr \ldots}$$

The radius of convergence is the radius of the largest disk around $0$ in which $1/(1+e^z)$ is analytic, i.e. the distance to the closest points where $1 + e^z = 0$. These points are $\pm i\pi$, so the radius of convergence is $\pi$.

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  • $\begingroup$ Thank you very much. I just finished doing all the derivatives (up to the fourth). I see why we need that radius, it makes sense since we are working with analytic functions. My coefficients match up exactly with yours and it seems as if the series contains every odd power of $z$ and each even power of $z$ is 0 except for when $n=0$ $\endgroup$ – H5159 Mar 25 '15 at 6:59
  • $\begingroup$ Yes: since $\dfrac{1}{1+e^z} = \dfrac{2}{1-e^{2z}} - \dfrac{1}{1-e^z}$, we have $a_n = (1-2^{n+1}) B_{n+1}/(n+1)!$ where $B_n$ are the Bernoulli numbers. $\endgroup$ – Robert Israel Mar 25 '15 at 15:50
  • $\begingroup$ Actually you don't need that to see that the even terms for $n \ne 0$ are $0$: just note that $\dfrac{1}{1+e^z} - \dfrac{1}{2} = \dfrac{e^{-z/2} - e^{z/2}}{2(e^{-z/2} + e^{z/2})}$ is an odd function. $\endgroup$ – Robert Israel Mar 25 '15 at 16:54
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One option is: Write $$\frac{1}{1 + e^z} \cdot (1 + e^z) = 1$$ and expand both of the factors on the l.h.s. in Taylor series" Since $e^z \sim \sum_{k = 0}^{\infty} \frac{1}{k!} z^k$, we have $$1 + e^z = 1 + \left(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + O(z^5)\right) = 2 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + O(z^5).$$

Then, expand the (t.b.d.) Taylor series of $\frac{1}{1 + e^z}$ as $$a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 + O(z^5).$$

The product of these two series is $$2 a_0 + (a_0 + 2 a_1) z + \left(\frac{1}{2}a_0 + a_1 + 2 a_2\right) + (\text{third- and fourth-order terms}) + O(z^5),$$ and the polynomial part here must agree with $1$ to fourth order. (Note that we only need consider the involved Taylor expansions here to order $4$ to determine $a_0, \ldots, a_4$, that is, to determine the desired Taylor series to that order.)

So, we get a system of five linear equations in the coefficients $a_0, \ldots, a_4$: \begin{align} 1 &= 2 a_0 \\ 0 &= a_0 + 2 a_1 \\ 0 &= \frac{1}{2} a_0 + a_1 + 2 a_2 \\ \vdots &= \vdots . \end{align} Solving and substituting gives the desired Taylor series.

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  • $\begingroup$ Where do you get the third line and fourth line? Are you using geometric series for the third line? $\endgroup$ – H5159 Mar 25 '15 at 6:34
  • $\begingroup$ No, every Taylor series (based at $z = 0$) has the form of the third line, $a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 + O(z^5)$. The fourth line is just the product of two series; computing it is just like computing the product of the given polynomial parts, and this only determines it (in this case) modulo terms of order $O(z^5)$. $\endgroup$ – Travis Willse Mar 25 '15 at 8:25
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To expand on my comment:$$\frac{1}{1+e^z}=\frac{1}{1-(-e^z)}=\sum_{n=0}^\infty (-e^z)^n$$ The series converges if $|-e^z|<1$.

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    $\begingroup$ But this isn't a Taylor series. $\endgroup$ – Robert Israel Mar 25 '15 at 6:30
  • $\begingroup$ Ah my mistake, I overlooked "Taylor" and went straight for series. $\endgroup$ – user170231 Mar 25 '15 at 6:31

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