Show $\displaystyle \lim_{x \to 0} \Bigl(\ln \sum_{k=0}^r x^k \Bigr)\biggl/\Bigl( \sum_{k=1}^{\infty}\frac{x^k}{k!}\Bigr) = 1.$

Question

Consider any finite integer $r \geq 2$. Then show $$\lim_{x \to 0} \dfrac{\ln\displaystyle \sum_{k=0}^r x^k }{\displaystyle \sum_{k=1}^{\infty}\frac{x^k}{k!}} = 1.$$

I can't understand how is it derived. I can see that denominator is $e^x-1$, and numerator is $\frac{1-x^{r+1}}{1-x}$

Correction: You missed a $\ln$ in the numerator. – Guy Mar 31 '14 at 05:29 — Guy, Mar 31 '14 at 05:29

score 4 · Accepted Answer · answered Mar 31 '14 at 05:29

4

Hint: Use L'Hospital's Rule. It works quite quickly.

answered Mar 31 '14 at 05:29

André Nicolas

507,029

YES! Got it! Also, thank you so much, for taking me to Star & Bars, it has helped me too much in different types of questions! It is so easy to learn & to apply! – Silent Mar 31 '14 at 05:50
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You are welcome. It is a useful method. I gave the Wikipedia reference for two reasons. (i) The article is well done and (ii) I have given explanations of Stars and Bars on MSE more than once, so have others, it seemed unreasonable to do it yet again. – André Nicolas Mar 31 '14 at 05:53

score 2 · Answer 2 · answered Mar 31 '14 at 05:29

2

$$ \log \sum_{k=0}^r x^k= \log(1-x^{r+1})-\log(1-x)=-x^{r+1}+o(x^{r+1})+x+o(x) = x+o(x) \\ e^x-1=x+o(x). $$Now the ratio gives the result.

answered Mar 31 '14 at 05:29

mookid

28,236

How did you get $\ln(1-x^{r+1})-\ln(1-x)=e^x-1$ Please explain! I will be obliged! – Silent Mar 31 '14 at 05:40
This is not what I get. What i get is $\ln(1−x^{r+1})−\ln(1−x)=x+x\epsilon(x)$ with $\lim\epsilon = 0$, and the same for $\exp x -1$ but for a different function $\epsilon$. – mookid Mar 31 '14 at 05:48
OK , still confused! Though thank you! – Silent Mar 31 '14 at 05:51

score 2 · Answer 3 · answered Mar 31 '14 at 05:49

The numerator of the expression is $$\log \left(\frac{1-x^{r+1}}{1-x}\right)$$ (not $r$ as you wrote in the post) and the denominator is $(e^x-1)$ as you noticed. You can expand the logarithm as the difference of two terms and apply L'Hospital's Rule as suggested by André Nicolas. This should work very quickly.

Another solution would be Taylor expansion as suggested by mookid taking into account the fact that, when $x$ goes to $0$, $x^{r+1}$ is negligible compared to $x$. So, you are basically let with the limit of $$-\frac{\log (1-x)}{x}$$ for which whatever method will take you to the result.

Show $\displaystyle \lim_{x \to 0} \Bigl(\ln \sum_{k=0}^r x^k \Bigr)\biggl/\Bigl( \sum_{k=1}^{\infty}\frac{x^k}{k!}\Bigr) = 1.$

3 Answers3