# A problem!

#### Also sprach Zarathustra

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

My question is:

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

What is form of $$\displaystyle a_n$$?

Thank you very much!

Last edited:

#### Ackbeet

MHF Hall of Honor
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?

#### Also sprach Zarathustra

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!

#### Ackbeet

MHF Hall of Honor
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.

#### Also sprach Zarathustra

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

#### Ackbeet

MHF Hall of Honor
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.

#### Also sprach Zarathustra

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).

#### simplependulum

MHF Hall of Honor
Hi , it is $$\displaystyle H_n = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n}$$

• Also sprach Zarathustra

#### Also sprach Zarathustra

Make sense!

How you get this result please?

#### Also sprach Zarathustra

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

Thanks!

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