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(Motivation) As homework, we have been asked to prove that the following series converges: $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}$$ I did it in two ways:

  1. Using the alternating series test (Leibniz criterion), proving that $\frac{n^3}{4^n}$ is decreasing and also $\lim_\limits{n\to +\infty}{\frac{n^3}{4^n}}=0$.

  2. Using the n-th root test (Cauchy's criterion) and absolute convergence, proving that $\lim_\limits{n\to +\infty}{\frac{(\sqrt[n]n)^3}{4}}=\frac{1}{4}<1$.

However, Wolfram Alpha states another interesting result: That this series sums exactly to $-0.0064$. I would like to see how that result is obtained, so a proof for it.

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As in other answers consider $$\sum_{n=1}^{+\infty}n^3 x^n$$ and rewrite $$n^3=n(n-1)(n-2)+a n(n-1)+bn$$ Expanding and identifying coefficients, you should get $a=3$ and $b=1$. So $$\sum_{n=1}^{+\infty}n^3 x^n=\sum_{n=1}^{+\infty}n(n-1)(n-2) x^n+3\sum_{n=1}^{+\infty}n(n-1) x^n+\sum_{n=1}^{+\infty}n x^n$$ $$\sum_{n=1}^{+\infty}n^3 x^n=x^3\sum_{n=1}^{+\infty}n(n-1)(n-2) x^{n-3}+3x^2\sum_{n=1}^{+\infty}n(n-1) x^{n-2}+x\sum_{n=1}^{+\infty}n x^{n-1}$$

$$\sum_{n=1}^{+\infty}n^3 x^n=x^3\left(\sum_{n=1}^{+\infty} x^{n}\right)'''+3x^2\left(\sum_{n=1}^{+\infty} x^{n}\right)''+x\left(\sum_{n=1}^{+\infty} x^{n}\right)'$$ Now, for $|x|<1$ $$\sum_{n=1}^{+\infty} x^{n}=\frac x {1-x}\qquad \left(\sum_{n=1}^{+\infty} x^{n}\right)'=\frac{1}{(1-x)^2}$$ $$\left(\sum_{n=1}^{+\infty} x^{n}\right)''=\frac{2}{(1-x)^3}\qquad \left(\sum_{n=1}^{+\infty} x^{n}\right)'''=\frac{6}{(1-x)^4}$$ which make $$\sum_{n=1}^{+\infty}n^3 x^n=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac{x}{(1-x)^2}={x(x^2+4x+1)\over (1-x)^4}$$ Using now $x=-\frac 14$ leads to $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-\frac{4}{625}=-0.0064$$

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Hint. By differentiating and multiplying by $x$ three times the standard geometric identity $$ \sum _{n=1}^{\infty }x^{n}=\frac1{1-x},\qquad |x|<1, $$ one gets $$ \sum _{n=1}^{\infty }n^3x^{n}=\frac{x^3+4x^2+x}{(1-x)^4},\qquad |x|<1, $$ then putting $x:=-\dfrac14$ gives the result.

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You're not yet probably at the point you can see this, but it comes from the fact that for $|x|<1$ we have that

$$\sum_{n=1}^\infty n^3x^n={x(x^2+4x+1)\over (1-x)^4}$$

Plugging in $x=-{1\over 4}$ you get the result.

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    $\begingroup$ @Jason each time you apply the operator $x\cdot{d\over dx}$, it multiplies the coefficient of $x^n$ by $n$. I just did it 3 times to ${1\over 1-x}$ $\endgroup$ – Adam Hughes Dec 5 '16 at 23:03
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You can get this by clever sum manipulations.

Start from $$ \sum \frac1{(-4)^n}=-\frac14+\frac1{16}-\frac1{64} + \cdots = -\frac15 $$ Then look at $$ \sum \frac{n}{(-4)^n}=\begin{array}{lllll} -\frac14&+\frac1{16}&-\frac1{64} &+ \cdots \\ &+\frac1{16}&-\frac1{64}&- \cdots \\ &&-\frac1{64}&+ \cdots \end{array}\\ =\sum \frac1{(-4)^n} - \frac14\sum \frac1{(-4)^n} + \frac1{16}\sum \frac1{(-4)^n} \cdots = -\frac45\cdot \frac{1}{5} = -\frac{4}{25} $$ Next do the same sort of vertical breakup, but with $\sum \frac{n}{(-4)^n}$ in each row: $$ \begin{array}{lllll} -\frac14&+\frac2{16}&-\frac3{64} &+\frac4{256} &-\cdots \\ &+\frac1{16}&-\frac2{64}&+\frac3{256} &-\cdots \\ &&-\frac1{64}&+\frac2{256} &- \cdots \\ &&&+\frac1{256} &-\cdots \end{array} = \sum\frac{n(n+1)}{2}\frac1{(-4)^n} \\ =\sum \frac{n}{(-4)^n} - \frac14\sum \frac{n}{(-4)^n} + \frac1{16}\sum \frac{n}{(-4)^n} \cdots = -\frac45\cdot \frac{4}{25} = -\frac{16}{125} $$ And the next step of the same trick, using rows of $\sum\frac{n(n+1)}{2}\frac1{(-4)^n}$, will give $$\sum\frac{n(n+1)(n+2)}{6}\frac1{(-4)^n} = -\frac45\cdot\frac{16}{125}=-\frac{64}{625} $$

Finally, write $$ n^3 = 6 \frac{n(n+1)(n+2)}{6} -6 \frac{n(n+1)}{2} + n $$ and you get the answer $$ -6\cdot\frac{64}{625} +6\cdot\frac{16}{125} -\frac{4}{25} =-0.0064 $$

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(1). If a power series $\sum_{n=0}^{\infty}A_nx^n$ converges for some $x\ne 0$ then $\sum_{n=0}^{\infty}A_ny^n$ converges for all real or complex $y$ such that $|y|<|x|.$ This follows from the Hadamard Radius Formula:

Let $S=\lim_{n\to \infty}\sup_{m> n}|A_m|^{1/m}.\;$ Let $R=1/S$ (with $R=\infty$ if $S=0$). The series converges when $|x|<R$ and diverges for $|x|>R.$

The proof is not difficult. (I recall a short, clear presentation of it in Complex Analysis, by Ahlfors.)

(2). If $|A_n|=n^3/4^n$ then, as $\lim_{n\to \infty}(n^3)^{1/n}=1,$ the series $\sum_nA_nx^n$ converges when $|x|<4.$ In particular, when $x=\pm 1.$

(3). When $f(x)=\sum_nA_nx^n$ for $|x|<R$ (with $0<R\leq \infty$), differentiating or integrating the terms of the series yields series for $f'(x)$ and $\int f(x)dx.$ This is useful, and significant because, in general, interchanging the order of limiting processes often gives unequal results.

(4). For $|x|<4:$

Let $f(x)=1-\sum_{n=1}^{\infty}(-1)^nn^3x^n/4^n $. We want to find $f(1)-1.$ Let $$g(x)=\sum_{n=0}^{\infty}(-1)^nx^n/4^n=\left(1+\frac {x}{4}\right)^{-1}.$$ The RHS above is easily repeatedly differentiated. Differentiating the LHS repeatedly and comparing the co-efficients to those of $f(x),$ we have $$(x/4)^3g'''(x)-6(x/4)^2g''(x)+7(x/4)g'(x)-g(x)=f(x),$$ from which $f(1)-1$ is found.

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