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Let us denote $S_{n}(x)=1+\frac{x}{1 !}+\frac{x^{2}}{2!}+ ... + \frac{x^{n}}{n!}$.

  1. How could be calculated the limit $$L(x)=\lim_{n\to \infty}\frac{S_{n}(n x)}{e^{n x}}=\lim_{n\to \infty}\frac{1+\frac{nx}{1 !}+\frac{(nx)^{2}}{2!}+ ... + \frac{(nx)^{n}}{n!}}{e^{n x}}, x\ge 0,\,\,\,\ ?$$

  2. Similar question, when $x\ge 0$ above is replaced by $z\in \mathbb{C}$ ?

  3. More general, if $r_{n}$ is a sequence of real numbers with $r^{n}\searrow 1$, what are the limits $\lim_{n\to \infty}\frac{S_{n}(n r_{n}x)}{e^{n x}}$ for $x\ge 0$ and $\lim_{n\to \infty}\frac{S_{n}(n r_{n}z)}{e^{n z}}$ for $z\in \mathbb{C}$ ? (Are they equal with the limits from the above points 1) and 2) ?

Initially, my intuition told me that probably that the limit $L(x)$ is equal to one, for any $x\ge 0$. But thinking better, my opinion is that the limit $L(x)=0$, for all $x\ge 0$. In support to this guess, for example, for $x=1$ I have calculated $\frac{S_{n}(n)}{e^{n}}$ for several consecutive values of $n$ and it appeared to me that it forms a decreasing sequence. In the general case, I have tried to use the Stolz-Cesaro lemma to the ratios $\frac{S_{n}(nx)}{e^{n x}}$ and $\frac{e^{nx}}{S_{n}(nx)}$, but it did not work. Also, I have tried to estimate $|S_{n}(nx)-e^{nx}|$ by using the Lagrange form of the remainder for Taylor series, but again I was not able to get any conclusion.In the complex case, the situation seems to be more intricated. Indeed, for $z=i$, we get $$\frac{S_{n}(n i)}{e^{n i}}=\frac{S^{(cos)}_{n}(n)+iS^{(sin)}_{n}(n)}{cos(n)+isin(n)},$$ where $S^{(cos)}_{n}$ and $S^{(sin)}_{n}$ represents the partial sums of order $n$ from the series development of cosine and sine functions. The limit with $n\to \infty$ in this case looks more tricky, as do not exist the limits $\lim_{n\to \infty}cos(n)$ and $\lim_{n\to \infty}sin(n)$.

Daniel Fischer
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1 Answers1

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This limit equals $0$ when $z$ is inside certain closed curve $\gamma$ and equals $\infty$ when $z$ is outside $\gamma$. This curve $\gamma$ is the boundary of the region $\{z:|z|<1,|ze^{1-z}|<1\}$, and it is called the Szego curve.

This follows from a result of Szego which gives a precise asymptotic behavior of the ratio $e^{-nz}S_n(nz)$.

G. Szego, Uber eine Eigenschaft der Exponentialreihe, Berlin Math. Ges. Sitzunsber. 23 (1924), 50-64.

If you have no access to this journal or do not read German, just type on Google: "Zeros of partial sum of the exponential" and you will see about 10 papers on this, most of them freely available in English.

Alexandre Eremenko
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  • Dear Professor Eremenko, Thank you so much for your very valuable answer and suggested references. ! – Sorin Gal Apr 06 '15 at 19:26
  • @Sorin Gal: Instead of writing your thanks, you can just up vote my answer: it is one click on the arrow on the right of it:-) – Alexandre Eremenko Apr 06 '15 at 22:38
  • I apologize, I did not know this rule, I clicked on "votes", is that correct now ? – Sorin Gal Apr 07 '15 at 14:58
  • By the way, I have in my mind some other related interesting questions (in my opinion !), (some coming from approximation theory). Would you agree that if these questions will still remain "suspended" after reading the references you have indicated, to contact you directly by email ? Best wishes, Sorin – Sorin Gal Apr 07 '15 at 15:09
  • @Sorin Gal: do you see the arrows on the LEFT of my answer? If you like the answer, click the arrow pointing up. (If you don't, click on the arrow pointing down). – Alexandre Eremenko Apr 07 '15 at 20:09
  • I did it, it is correct now ? – Sorin Gal Apr 07 '15 at 20:26
  • Yes. If 1 appears instead of 0 between the arrows, you did it correctly. Always do this for questions or answers which you like or do not like. – Alexandre Eremenko Apr 07 '15 at 20:32