1. ## term-by-term differentiation

Find the sum using term-by-term differentiation:

$f(x) = \frac{x^3}{1\cdot 3} + \frac{x^4}{2\cdot 4} + \frac{x^5}{3\cdot 5} + \frac{x^6}{4\cdot 6}...$

I just don't know how to deal with the $'n'$ term. Taking the third derivative I get, for $n \in [3,\infty), (n-1)x^{n-3} = \frac{n-1}{1-x} = ...?$

2. Originally Posted by Anonymous1
Find the sum using term-by-term differentiation:

$f(x) = \frac{x^3}{1\cdot 3} + \frac{x^4}{2\cdot 4} + \frac{x^5}{3\cdot 5} + \frac{x^6}{4\cdot 6}...$

I just don't know how to deal with the $'n'$ term. Taking the third derivative I get, for $n \in [3,\infty), (n-1)x^{n-3} = \frac{n-1}{1-x} = ...?$
Here's my thought:

$f'(x)=x\sum_{n=1}^{\infty}\frac{x^n}{n}=x\sum_{n=0 }^{\infty}\frac{x^{n+1}}{n+1}$

The derivative of the sum (not including the $x$ outside it) is $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$

So the sum itself is $\int\frac{1}{1-x}\,dx=-\ln(1-x)$ and it follows that $f'(x)=-x\ln(1-x)$.

Integrate that to find $f(x)$.

Note that $f(x)$ will only be defined on $[-1,1)$.

3. Originally Posted by Anonymous1
Find the sum using term-by-term differentiation:

$f(x) = \frac{x^3}{1\cdot 3} + \frac{x^4}{2\cdot 4} + \frac{x^5}{3\cdot 5} + \frac{x^6}{4\cdot 6}...$

I just don't know how to deal with the $'n'$ term. Taking the third derivative I get, for $n \in [3,\infty), (n-1)x^{n-3} = \frac{n-1}{1-x} = ...?$
It is not hard to see that
$f(x)=\sum_{k=1}^{\infty}\frac{x^{k+2}}{k(k+2)}$.
Here we have $a_{k}:=\frac{1}{k(k+2)}$ for all $k\in\mathbb{N}$.
For this power series, we see that the radius of convergence is $r:=\lim_{k\to\infty}\frac{a_{k}}{a_{k+1}}=\lim_{k\ to\infty}\frac{k(k+2)}{(k+1)(k+3)}=1$.
Hence $\sum_{k=1}^{\infty}\frac{x^{k+2}}{k(k+2)}$ converges for all $|x|, you may check the end points.
Thus, this power series converges uniformly in $[-r_{0},r_{0}]$ for any $r_{0}\in[0,r)=[0,1)$, hence term-by-term differentiable (and integrable) on the same interval.
As we may pick $r_{0}$ arbitrarily closer to $r=1$, we get
$f^{\prime}(x)=\bigg(\sum_{k=1}^{\infty}\frac{x^{k+ 2}}{k(k+2)}\bigg)^{\prime}$
........ $=\sum_{k=1}^{\infty}\bigg(\frac{x^{k+2}}{k(k+2)}\b igg)^{\prime}$
........ $=\sum_{k=1}^{\infty}\frac{(k+2)x^{k+1}}{k(k+2)}$
........ $=\sum_{k=1}^{\infty}\frac{x^{k+1}}{k}$
for all $|x|.

See Uniform Convergence -- from Wolfram MathWorld