Fractions with Factorials

My solution is different. $S = \sum _{n=2} ^\infty {\frac {n-1}{n!}} = \sum _{n=2} ^\infty {\left( \frac {n}{n!} - \frac {1}{n!} \right)} = \sum _{n=2} ^\infty {\left( \frac {1}{(n-1)!} - \frac {1}{n!} \right)}$ $= \frac {1}{1!} - \frac {1}{2!} + \frac {1}{2!} - \frac {1}{3!} + \frac {1}{3!} - \frac {1}{4!}... = \boxed {1}$

We may use taylor series expansion of $e^{x}$

$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...$

Dividing throughout by x,

$\frac{e^{x}}{x}=\frac{1}{x}+1+\frac{x}{2!}+\frac{x^{2}}{3!}+\frac{x^{3}}{4!}+...$

Differentiate both sides wrt x,

$\frac{xe^{x}-e^{x}}{x^{2}}=\frac{-1}{x^{2}}+\frac{1}{2!}+\frac{2x}{3!}+\frac{3x^{2}}{4!}+...$

Substitute $x=1$ ,

$0=-1+\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...$

So the sum of given series is $\boxed{1}$

Recall that $\displaystyle \sum_{k=0}^\infty \frac {1}{k!} = e$

Hence,

$\begin{aligned} \displaystyle \frac {1}{2!} + \frac {2}{3!} + \frac {3}{4!} + \frac {4}{5!} + \ldots & = & \sum_{k=0}^\infty \frac {k+1}{(k+2)!} \\ & = & \sum_{k=0}^\infty \frac {k+2-1}{(k+2)!} \\ & = & \sum_{k=0}^\infty \frac {k+2}{(k+2)!} - \sum_{k=0}^\infty \frac {1}{(k+2)!} \\ & = & \sum_{k=0}^\infty \frac {1}{(k+1)!} - \sum_{k=0}^\infty \frac {1}{(k+2)!}, \text{telescoping series} \\ & = & \frac {1}{(0+1)!} = \boxed{1} \\ \end{aligned}$

(2-1)/2! +(3-1)/3!+(4-1)/4!+.....= 1

$\displaystyle S\quad =\quad \sum _{ n=1 }^{ \infty }{ \frac { n }{ (n+1)! } }$

$\displaystyle S\quad +\quad (e-2)\quad =\quad \sum _{ n=1 }^{ \infty }{ \frac { n }{ (n+1)! } } \quad +\quad \sum _{ n= 1 }^{ \infty }{ \frac { 1 }{ (n+1)! } }$

$\displaystyle S\quad +\quad (e-2)\quad =\quad \sum _{ n=1 }^{ \infty }{ \frac { (n+1) }{ (n+1)! } }$

$\displaystyle S\quad +\quad (e-2)\quad =\quad e-1$

$\displaystyle S\quad =\quad 1$

If we have forgotten about the Taylor series expansion of $e$ , it suffices to consider the partial sums of $\sum_{i=1}^{\infty} \frac{i}{(i+1)!}$ .

Notice that $S_n = \sum_{i=1}^{n} \frac{i}{(i+1)!} = 1 - \frac{1}{(n+1)!}$ (Proof by induction)

Sketch of Proof by Induction: $H_n: S_n = 1 - \frac{1}{(n+1)!}$

1. Verify $H_1: S_1 = \frac{1}{2}$ is true.

2. Assume for some $n$ that $H_n$ is true, and consider $H_{n+1}$ ,

$\begin{aligned} \left(\sum_{i=1}^{n} \frac{i}{(i+1)!}\right) + \frac{n+1}{(n+2)!} & = \left(1 - \frac{1}{(n+1)!}\right) + \frac{n+1}{(n+2)!} \\ & = 1 - \frac{(n+1)(n+1)! - (n+2)(n+1)!}{(n+1)!(n+2)!} \\ & = 1 - \frac{n+1}{(n+2)!} \end{aligned}$

By the principle of induction, $H_n$ is true for all $n$

Then it is straightforward to see that $\lim_{n\rightarrow \infty} S_n = \boxed{1}$

dumb ways as always :p

The first term of the series is $\frac{1}{2!} = \frac{1}{2}$ since the remainder of the series gets increasingly less than that, it is only logical to assume that the answer will end up being the closest integer away from zero (this is because the series only adds numbers so the answer must be above the first term). This makes the answer $1$ logically.

Note - The above method doesn't always work.

Adding

1/2! + 1/3! + 1/4 ! + ...

You will get

2/2! + 3/3! + 4/4! + ...

Which equal to 1 + 1/2! + 1/3! + 1/4! + ,,,

Now substract 1/2! + 1/3! + 1/4 ! + ...

So the answer is 1

the first term is 1/2 and then it goes on decreasing and since it is an integer 1 or 2 is the answer ....by hit and trial

Put s = 1/2! + 2/3! + 3/4! + .....

(Assume s converges)

Now, e = 1 + 1/1! + 1/2! + 1/3!+....

 =>     s + e = 1 + 1/1! + 2/2! + 3/3! + 4/4! + ...
                 = 1 + [1 + 1/1! + 1/2! + 1/3! + ...]
                 = 1 + e

  =>s = 1