1. ## A problem!

$\displaystyle f(x)=x-\frac{3}{2}x^2+\frac{11}{6}x^3-\frac{50}{24}x^4+\frac{274}{120}x^5+...$
This is Taylor series of function $\displaystyle f(x)=\frac{ln(1+x)}{1+x}$

My question is:

$\displaystyle f(x)=x-\frac{3}{2}x^2+\frac{11}{6}x^3-\frac{50}{24}x^4+\frac{274}{120}x^5+...=\Sigma^\in fty_{n=0}(-1)^{n+1}a_n x^n$

What is form of $\displaystyle a_n$?

Thank you very much!

2. Well, the denominators look something like a factorial (except for that pesky 3 in the squared term; perhaps you could massage it to look better). Maybe you could set up a recurrence relation to figure out the numerators?

3. Ohhh... it's not 2/3 it's 3/2

Sorry!!!

(I'm going to change it in my original post)

Thank you Mr. Keister!

4. Aha. So it is a nice little factorial in the denominator. Well, the recurrence relation idea sounds nice, but I'm not sure how it would work in principle. Why don't you show all the details of the computations for those first few terms? Maybe something'll pop out.

5. Originally Posted by Ackbeet
Aha. So it is a nice little factorial in the denominator. Well, the recurrence relation idea sounds nice, but I'm not sure how it would work in principle. Why don't you show all the details of the computations for those first few terms? Maybe something'll pop out.

Here it is: Function calculator

Moreover... look here...

1 3 11 50 - OEIS Search Results

6. No, that's not what I'm interested in. I'm interested in seeing the step-by-step calculations for computing each coefficient. I'm curious to see if a pattern pops out by examining that process.

7. Unfortunately, I can't do so, it is very complicated, but the thing I can do is to post the original question asked.

Q.
Find Taylor polynomial of $\displaystyle f(x)=\frac{ln(1+x)}{1+x}$ around $\displaystyle x=0$, find formula for $\displaystyle f^{(n)}(x)$.
Find the radius of convergence of infinite Taylor polynomial(power series).

8. Hi , it is $\displaystyle H_n = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n}$

9. Make sense!

How you get this result please?

10. So, the radius of convergence $\displaystyle R$ of $\displaystyle \Sigma^{\infty}_{n=0}(-1)^nH_nx^n$ is R=0?

Thanks!

11. No , the radius is 1 , i obtain the general term by expanding $\displaystyle \ln(1+x)$ and $\displaystyle \frac{1}{1+x}$ followed by coefficients' consideration .

12. The general formula for the product of two series is...

$\displaystyle \displaystyle \sum_{n=0}^{\infty} a_{n}\ x^{n}\ \sum_{k=0}^{\infty} b_{k}\ x^{k} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}\ b_{n-k}\ x^{n}$ (1)

... and in Your case is...

$\displaystyle \displaystyle \ln (1+x)= \sum_{n=1}^{\infty} (-1)^{n+1}\ \frac{x^{n}}{n}$

$\displaystyle \displaystyle \frac{1}{1+x} = \sum_{k=0}^{\infty} (-1)^{k}\ x^{k}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

13. I have difficulties in finding radius of convergence! Why it equals to 1?

In other words, WHY $\displaystyle R=\lim_{n\to\infty}\frac{1}{\sqrt[n]{H_n}} = 1$?

14. Maybe it from the "sandwich" rule?
$\displaystyle 1<H_n<n$
$\displaystyle \sqrt[n]{1}<\sqrt[n]H_n<\sqrt[n]{n}$

so when $\displaystyle n\to\infty$

$\displaystyle H_n \to 1$

15. If $\displaystyle \lim_{n \to \infty} \frac{a_{n + 1}}{a_n}$ exists, then $\displaystyle \lim_{n \to \infty} \sqrt[n]{a_n}$ does as well, and they are equal. Since the first expression exists, you can use it to find the radius of convergence, which will work out to be 1.

If you think to calculate $\displaystyle a_n - a_{n - 1} = \frac{1}{n}$ from the terms you listed out then it becomes pretty easy to figure out the form of the $\displaystyle a_n$, though this isn't exactly a rigorous solution.