Solving the series

Calculus Level 3

Recall that
$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.$ Then what is the value of $\large \frac{\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots}{\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \cdots}\, ?$

Notation: $!$ is the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

\frac{e^2}{2}

\frac{1}{e}

\frac{1}{e^2}

\frac{2}{e^3}

3 solutions

Naren Bhandari
Feb 18, 2017

Evaluating the numerator as :- $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots$

$\frac{3-1}{3!} + \frac{5-1}{5!} + \frac{7-1}{7!} + \cdots$

$\frac{3}{3!} - \frac{1}{3!} + \frac{5}{5!} - \frac{1}{5!} + \frac{7}{7!} - \frac{1}{7!}+ \cdots$

$\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} - \frac{1}{7!} +\cdots$

$1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} - \frac{1}{7!} +\cdots$

$= e^{-1}$

Since $e^{-1} = 1 - \frac{1}{1!}+\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} - \frac{1}{7!} +\cdots$

Same as Evaluting the denominator as

$\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \cdots$

$\frac{1+1}{1!} + \frac{3+1}{3!} + \frac{5+1}{5!} + \cdots$

$1+ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} +\frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \cdots$

$=e$

Therefore,

$\frac{e^{-1}}{e} = \boxed{\frac{1}{e^2}}$

Moderator note:

The solution uses the "trick" that $\frac{ n } { n!}$ simplifies to $\frac{1}{(n-1)!}$ .

In a similar vein, can you evaluate $\sum_{i=1}^n i \times i !$ as a telescoping sum?

Nice solution! I like how you split each fraction as sum/difference of two fractions. What motivated you to split the fractions the way you did?

Pranshu Gaba - 4 years, 3 months ago

$S_n = \sum_{i=1}^N i\times i! = 1.1! + 2.2! + 3.3! + 4.4! + 5.5! + 6.6! +\cdots N.N!$

$S_n = \sum_{i=1}^N = 1 + 4 + 18 + 96 + 600 + 4370 + \cdots N.N!$

$S_n = (2-1) + (6-2) + (24-6) + (120-24) + (720-120) + (5040 -720) +\cdots + (n+1)! -(n-1)!$

$S_n = (2!-1!) +(3! -2!) + (4! -3!) + (5! -4!) + (6!-5!) + (7!-6!) + \cdots + \left((n+1)! - (n-1)! \right)$

$S_n = -1! + (2! -2!) +(3! -3!) + (4! -4!) + (5!-5!) +\cdots +( n-1)! - (n-1)! + (n+1)!$

Since all the terms get cancelled except $-1$ and $(n+1)!$ which shows that $\sum_{i=1}^N i\times i$ Is a telescoping series which is equal to

$S_n = (n+1)! -1$

Naren Bhandari - 4 years, 3 months ago

Last term should be ((n+1)! - (n)!) so that it cancel out with n! from the previous term. It shouldn't be ((n+1)! - (n-1)!).

Also typo in the first line in the solution for original question, (i+1)! = (i+1)(i!)

Bharath Kanigiri - 3 years, 2 months ago

@Pi Han Goh Thank you , Sir :)

Naren Bhandari - 4 years, 3 months ago

We can see that

$(i+1)! = i! (i+1)! =i\times i! + i!$

Then

$i\times i! = (i+1)! - i!$

And we have that

$\sum_{i=1}^n i\times i! =\sum_{i=1}^n ((i+1)! - i!)$

This is a telescopic sum, that is, only the values $-1!$ and $(n+1)!$ remain, the rest cancel out. So in the end

$\sum_{i=1}^n i\times i! = (n+1)! - 1$

Ana Echavarria - 4 years, 3 months ago

Yup, that's just telescoping sum in disguise! Well done~~

Pi Han Goh - 4 years, 3 months ago

There's a typing mistake in the first line.

Atomsky Jahid - 4 years, 2 months ago

Did the same as Naren's solution.

Rohit Vijayvargiya - 3 years, 4 months ago

Thank you for the compliment :) Well, there is no such any motivation however for the numerator terms and denominator i just tried to expend so that i could get exact match with $e^x$ series. Moreover, the term in numerator are the pattern of $\frac{n}{n+1)!}$ and better spliting the factorial part i splited without factoriart part . :)