Sum of power series using only differentiation

Question

Find the sum of $\sum _{n=1}^\infty n (2 n+1) x^{2 n}$ using only differentiation and knowing that $\sum _{n=0}^\infty x^n=\frac{1}{1-x}$

I started like that:

$n^2=n(n-1)+n$

so $\sum _{n=1}^\infty n(2n+1)x^{2n}=\sum _{n=1}^\infty (2n^2+n)x^{2n}=2\sum _{n=?}^\infty n(n-1)x^{2n}+3\sum _{n=?}^\infty nx^{2n}$

but as you can see I don't know at which $n$ does the summing start (that's why I marked it with "?").

Then I would go $x^2=t$

so $2\sum _{n=?}^\infty n(n-1)t^{n}+3\sum _{n=?}^\infty nt^{n} = 2t^2\sum _{n=?}^\infty n(n-1)t^{n-2}+3t\sum _{n=?}^\infty nt^{n-1} = 2t^2(\sum _{n=0?}^\infty t^n)''+3t(\sum _{n=0?}^\infty t^n)'=2t^2(\frac{1}{1-t})''+3t(\frac{1}{1-t})'$

and then it is simple.

But am I right? Can one calculate the sum like that? If no, how to find constant when integrating $\sum _{n=1}^\infty n(2n+1)x^{2n}$?

Because $\int \sum _{n=1}^\infty n(2n+1)x^{2n}dx=\sum _{n=1}^\infty n(2n+1)\frac{x^{2n+1}}{2n+1}+C$ and I don't know how to find the C if I don't know what function does represent our sum.

Answer 1

From $$\frac{1}{1 - x} = \sum^{\infty}_{n = 0}x^n$$

we make a simple substitution to deduce that:

$$\frac{1}{1 - x^2} = \sum^{\infty}_{n = 0}x^{2n}$$

Multiplying both sides with $x$,

$$\frac{x}{1 - x^2} = \sum^{\infty}_{n = 0}x^{2n + 1}$$

Differentiating both sides with respect to $x$ twice, we have:

$$\frac{d^2}{dx^2}\left(\frac{x}{1 - x^2}\right) = \sum^{\infty}_{n = 0}2n(2n + 1)x^{2n - 1}$$

Multiplying both sides by $\frac{x}{2}$,

$$\frac{x}{2}\frac{d^2}{dx^2}\left(\frac{x}{1 - x^2}\right) = \sum^{\infty}_{n = 0}n(2n + 1)x^{2n}$$

Answer 2

We have $\displaystyle\sum_{n=0}^\infty x^{2n}=\sum_{n=0}^\infty (x^2)^n=\frac1{1-x^2}$ for $|x^2|<1$

$$\implies \sum_{n=0}^\infty x^{2n+1}=\frac x{1-x^2}$$

$$\implies2\sum_{n=0}^\infty x^{2n+1}=\frac1{1-x}-\frac1{1+x}$$

Differentiating wrt $x$

$$\implies2\sum_{n=0}^\infty (2n+1)x^{2n}=\frac1{(1-x)^2}+\frac1{(1+x)^2}$$

Again differentiate wrt $x$

Answer 3

Observe that by differentiating $f(x)=\sum_{n=0}^{\infty} x^n$ once, you will form $$f'(x)=\sum_{n=0}^{\infty}nx^{n-1}=\sum_{n=1}^{\infty}nx^{n-1}.$$

Now, substituting $x$ by $x^2$ and multiplying by $x^3$ will make exponents $2n+1$ appear, giving $$x^3f'(x^2)=\sum_{n=1}^{\infty}nx^{2n+1}.$$

Derive once again to get $$(x^3f'(x^2))'=\sum_{n=1}^{\infty}n(2n+1)x^{2n}.$$ By chance, both the coefficients and exponents are now as desired.

To manipulate entire series, there are several rules you can use:

1. Differentiate: $f'(x)=\sum na_nx^{n-1}$. Multiplies the terms by the exponents.
2. Integrate: $\int f(x) dx=\sum \frac{a_n}nx^{n+1}$. Divides the terms by the exponents.
3. Raise $x$ to a power: $f(x^p)=\sum a_nx^{pn}$. Scales the exponents.
4. Multiply or divide by a power: $x^{\pm q}f(x)=\sum a_nx^{n\pm q}$. Shifts the exponents.

Answer 4

This is a homework problem, so I will only give good hints instead of working it outright.

Hint 1: You can write $n(2n+1)x^{2n} = \frac{x}{2}(2n+1)(2n)x^{2n-1}=\frac{x}{2}(x^{2n+1})''$

Hint 2: You can write $\sum_k x^{ak+b} = x^b\sum_k(x^a)^k$

Hint 3: You can move terms into the sum: $\sum_{k=N}^{\infty}c_k=\left(\sum_{k=0}^{\infty}c_k\right) -c_0-c_1-\cdots-c_{N-1}$

Answer 5

$$S=\sum _{n=1}^\infty n (2 n+1) x^{2 n}$$ $$2S=\sum _{n=1}^\infty 2n (2 n+1) x^{2 n}$$ $$\dfrac {2S}x=\sum _{n=1}^\infty 2n (2 n+1) x^{2 n-1}$$ So that: $$\dfrac {2S}x=\left(\sum _{n=1}^\infty x^{2 n+1}\right)''=\left(\dfrac x{1-x^2}\right)''$$ Finally: $$\implies S=\left(\dfrac {x^2}{1-x^2}\right)$$

Answer 6

Recall that $$n(2n+1)x^{2n}=\frac{1}{2}x\frac{d^2}{dx^2}(x^{2n+1}).$$

Then, $$ \sum_{n=0}^\infty n(2n+1)x^{2n}=\frac{1}{2}x\frac{d^2}{dx^2}\left(\sum_{n=0}^\infty x^{2n+1}\right)=\frac{1}{2}x\frac{d^2}{dx^2}\left(\frac{x}{1-x^2}\right). $$ yields from the fact that the series $\displaystyle \sum_{n=0}^\infty x^{2n+1}$ is uniformly convergent for values of $|x|<1$.

Next, with the aid of the formula $$ \frac{x}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}-\frac{1}{1+x}\right) $$ it readily follows $\displaystyle \frac{d^2}{dx^2}\left(\frac{1}{1-x}\right)=\frac{2}{(1-x)^3}$ and $\displaystyle \frac{d^2}{dx^2}\left(\frac{1}{1+x}\right)=\frac{2}{(1+x)^3}$ so that $$ \sum_{n=0}^\infty n(2n+1)x^{2n}=\frac{1}{2}x\frac{d^2}{dx^2}\left(\dfrac{x}{1-x^2}\right)=\dfrac{x}{(1-x)^3}-\frac{x}{(1+x)^3}.$$

Finally, by a straightforwardly simplification based on the binomial formula one can moreover show that the right-hand side of the previous formula is equal to $$ \frac{x}{(1-x)^3}-\frac{x}{(1+x)^3}=\frac{x((1+x)^3-(1-x)^3)}{(1-x^2)^3}=\frac{6x^2+2x^4}{1-3x^2+3x^4-x^6}.$$