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I have the following question:

Evaluate the infinite series: $$S=1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$$ (a) $\displaystyle\frac1e$ (b) $\displaystyle\frac{-1}e$ (c) $\displaystyle\frac{2}e$ (d) $\displaystyle\frac{-2}e$

Now in the book they have given a strange explanation using $$(n+1)^3=[n(n-1)(n-2)+6n(n-1)+7n+1].$$ I don't understand how did they get this. So I have two doubts:

  1. How did they get this "trick"?
  2. If you would have got this question in a competitive exam, how would you have solved it (in the minimum amount of time), either directly or with numerical methods?
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    $\begingroup$ The trick seems to be typical when you want to see $An(n-1)(n-2)+Bn(n-1)+Cn+D$. Just identify. $\endgroup$ Sep 5, 2015 at 14:54

5 Answers 5

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You want to evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)^3}{n!}$.

Now the idea (at the latest if you see some $e$ as a possible answer) should be to relate this to the well-known formula $$\sum_{n=0}^{\infty} \frac{x^n}{n!}=e^x$$ In this particular case (given the $(-1)^n$ included) it is useful to plug in $x=-1$ and obtain $$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}=e^{-1}=\frac{1}{e}$$ But you still have a $(n+1)^3$ in the numerator where you wish to have $1$.

So you might want to split this into several parts such that you get one (or several) related sums.

As an example of this technique, if you want to know $\sum_{n=0}^{\infty} \frac{n+1}{n!}$ you might try: $$\sum_{n=0}^{\infty} \frac{n+1}{n!}=\sum_{n=0}^{\infty} \frac{n}{n!}+\sum_{n=0}^{\infty} \frac{1}{n!}$$ Now, the second sum is just $e$ and the first one is (after cancelling) $$\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1}{n!}=e$$ and so the total sum is $2e$. Can you now understand the "trick" and tackle your problem by yourself?

Concerning the second question:

If one happens to know this "trick", it is not hard to get the result by just a few computations so it would be my preferential choice.

But if you sit in the exam and have absolutely no idea of how to solve this algebraically (and especially if you don't have to give a proof of your result!) it might be a good idea to estimate the sum by computing a few terms. At least the sign of the result ($\pm$) should be approachable...

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    $\begingroup$ Yes thanks a lot, Now I understood the trick. But can you please answer the 2nd part of the question? $\endgroup$
    – Kartik
    Sep 5, 2015 at 14:59
  • $\begingroup$ @Kartik by learning the trick now.Its a very common question :-)!! I felt the same as you the first time...! $\endgroup$
    – user220382
    Sep 5, 2015 at 15:29
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    $\begingroup$ I changed "nominator" into "numerator $\endgroup$
    – imranfat
    Sep 5, 2015 at 15:37
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In general, if $p_i(x)$ is a sequence of polynomials with $\deg p_i=i$, then it is relatively easy to prove that every polynomial can be written as a linear combination of a finite number of $p_i$. When the $p_i$ are monic, you can do this easily. Let $P(x)=a_kx^k + ... a_0$. Then $P(x)-a_kp_k(x)$ is a polynomial of lower degree. Find it's first coefficient of that, and proceed inductively.

The standard basis for polynomials is $e_i(x)=x^i$.

So in your case; $p_i(n)=x(x-1)\cdots(x-(i-1))$ with $p_0(x)=1$. We chose this particular basis because $$\frac{p_i(n)}{n!} =\begin{cases}\frac{1}{(n-i)!} & n\geq i\\0&0\leq n<i\end{cases}$$ So:$$\sum_{n=0}^\infty \frac{p_i(n)z^n}{n!} = z^ie^{z}$$

In your case, $z=-1$.

With these polynomials, we get: $$\begin{align}(n+1)^3 &= n^3+3n^2+3n+1\\ (n+1)^3-p_3(n) &= n^3 +3n^2+3n+1 - (n^3-3n^2+2n)\\& = 6n^2+n+1\\ (n+1)^3-p_3(n)-6p_2(n) &= 6n^2+n+1 - 6n(n-1)\\&= 7n+1\\ \end{align}$$

So you get: $$(n+1)^3=p_3(n) + 6p_2(n) + 7p_1(n) +p_0(n)$$

The key is that the leading coefficient of the cubic $(n+1)^3$ is $1$, so we subtract $1\cdot p_3(n)$. Then the leading coefficient of the remaining quadratic is $6$, so we subtract $6p_2(n)$. The leading coefficient of the remaining linear polynomial is $7$, so we subtract $7p_1(x)$. You are left with a constant.

The relationship between the standard basis of the polynomials and the basis $p_i(x)=x(x-1)\cdots(x-(i-1))$ is actually fairly interesting and comes up again and again. The $p_i(x)$ are often called "falling factorials."

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My way to solve it:

If the numerators were all $1$, we would have the well-known series for $e^{-1}$, obtained from the entire series

$$e^{-x}=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}.$$

Now we want to make numerators $(k+1)^3$ appear. Multiply by $x$ and derive, to get

$$(xe^{-x})'=\left(\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k!}\right)'=\sum_{k=0}^\infty(-1)^k\frac{(k+1)x^k}{k!}.$$

Repeating this three times,

$$(x(x(xe^{-x})')')'=\sum_{k=0}^\infty(-1)^k\frac{(k+1)^3x^k}{k!}.$$

Then evaluate at $x=1$.

$$e^{-x}\to (1-x)e^{-x}\to (1-3x+x^2)e^{-x}\to (1-7x+6x^2-x^3)e^{-x}\to -\frac1e.$$


For such problems, you can often start from a known Taylor series (in particular $e^x$ and $1/(1-x)$)and perform the following operations

  • multiply or divide by a power of $x$ (shifts the exponents),
  • derive on $x$ (multiplies by the exponents),
  • integrate on $x$ (divides by the exponents),

to modify the coefficients.

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Since $(n+1)^3 = \color{red}{1}\cdot n(n-1)(n-2)+\color{red}{6}\cdot n(n-1) +\color{red}{7}\cdot n+\color{red}{1}$ we have:

$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{(-1)^n (n+1)^3}{n!} &=& \frac{13}{2}+\sum_{n\geq 3}\frac{(-1)^n (n+1)^3}{n!}\\&=&\frac{13}{2}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-3)!}+\color{red}{6}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-2)!}+\color{red}{7}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-1)!}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\frac{13}{2}-\sum_{n\geq 0}\frac{(-1)^n}{n!}+6\cdot\sum_{n\geq 1}\frac{(-1)^n}{n!}-7\cdot\sum_{n\geq 2}\frac{(-1)^n}{n!}+\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\sum_{n\geq 0}(-1+6-7+1)\frac{(-1)^n}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n}{n!}=\color{red}{-\frac{1}{e}}.\end{eqnarray*}$$

We simply get the initial decomposition by induction. Obviously $(n+1)^3-n(n-1)(n-2)$ is a quadratic polynomial in $n$, whose leading term is $6n^2$. So we substract $6n(n-1)$ and we are left with a linear polynomial and so on. Another chance is to tackle our problem this way: $$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n+1)^3}{n!}&=&\sum_{n\geq 1}\frac{(-1)^n n(n+1)^2}{n!}+\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}\\&=&\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}-\sum_{n\geq 0}\frac{(-1)^n(n+2)^2}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n(2n+3)}{n!}\\&=&-3\sum_{n\geq 0}\frac{(-1)^n}{n!}+2\sum_{n\geq 0}\frac{(-1)^n}{n!}=\frac{2-3}{e}.\end{eqnarray*}$$

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Looking at the given identity: \begin{align} (n+1)^3 &= [n(n-1)(n-2)+6n(n-1)+7n+1] \\ &= n(n^2 - 3n + 2) + 6 n^2 + n + 1 \\ &= n^3 + 3 n^2 + 3 n + 1 = (n+1)^3 \end{align}

Now, consider differentiation of the exponential series: \begin{align} D \, e^{t} &= \sum_{n=0}^{\infty} \frac{1}{n!} \, \frac{d}{dt} \, t^{n} = \sum_{n=0}^{\infty} \frac{n \, t^{n-1}}{n!} \end{align} Further differentiation can be applied.

Now, consider the series in question: \begin{align} S &= 1 - \frac{2^{3}}{1!} + \frac{3^{3}}{2!} - \cdots \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3}}{n!} \\ &= \left. \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3} \, t^{n}}{n!} \, \right|_{t=1} \\ &= \left. t^{3} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1)(n-2) \, t^{n-3}}{n!} + 6 t^{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1) \, t^{n-2}}{n!} + 7 t \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n \, t^{n-1}}{n!} + e^{-t} \, \right|_{t=1} \\ &= \left. t^3 \, D^{3} e^{-t} + 6 \, t^{2} \, D^{2} e^{-t} + 7 \, t \, D e^{-t} + e^{-t} \right|_{t=1} \\ &= \left. (-t^3 + 6 \, t^2 - 7 \, t + 1) \, e^{-t} \right|_{t=1} \\ &= - \frac{1}{e} \end{align}

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  • $\begingroup$ Thanks for the answer but this explanation is already given in the book which I am solving. I just wanted the explanation of the "Trick" in the first identity, $\endgroup$
    – Kartik
    Sep 5, 2015 at 15:16

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