$\begin{aligned} S & = \sum_{n=1}^\infty \frac n{(n+1)!} \\ & = \sum_{n=1}^\infty \frac {(n+1)-1}{(n+1)!} \\ & = \sum_{n=1}^\infty \frac 1{n!} - \sum_{\color{#3D99F6}n=1}^\infty \frac 1{(n+1)!} & \small \color{#3D99F6} \text{Note that both } \sum_{n=1}^\infty \frac 1{n!} \text{ and } \sum_{n=1}^\infty \frac 1{(n+1)!} \text{ converge.} \\ & = \sum_{n=1}^\infty \frac 1{n!} - \sum_{\color{#D61F06}n=2}^\infty \frac 1{n!} \\ & = \frac 1{1!} \\ & = \boxed{1} \end{aligned}$

$\begin{aligned} \large\sum_{n=1}^\infty \frac{n}{(n+1)!}&=\large\sum_{n=1}^\infty \frac{(n+1)-1}{(n+1)!}\\ &=\large\sum_{n=1}^\infty \frac{1}{n!}-\sum_{n=1}^\infty \frac{1}{(n+1)!}\\ &=\large (e-1)-(e-2)\\ &=\large\boxed 1. \end{aligned}$

Recall the series of $e$ $e = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dfrac{1}{5!} +\cdots$ Now note down the given infinite series $S= \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dfrac{4}{5!}+ \dfrac{5}{6!} +\dfrac{6}{7!}+ \cdots$

Can the given series be formed like that the expansion of $e$ ? Yes, the thing is to remember is not to break(split) denominator part. . $S= \dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+\dfrac{5-1}{5!}+\dfrac{6-1}{6!} + \cdots \\ S = \left(\dfrac{2}{2!} + \dfrac{3}{3!} + \dfrac{4}{4!}+ \cdots\right) -\left(\dfrac{1}{2!}+ \dfrac{1}{3!} + \dfrac{1}{4!} + \cdots \right) \\ S= \left(\dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!}+\cdots\right)- \left( {\color{#3D99F6}1 + \dfrac{1}{1!}} + \dfrac{1}{1!} + \dfrac{1}{2!}+ \dfrac{1}{3!} + \cdots -{\color{#3D99F6}-1-\dfrac{1}{1!}}\right) \\ S = \left({\color{#3D99F6}1} + \frac{1}{1!} +\dfrac{1}{2!} + \dfrac{1}{3!} +\cdots {\color{#3D99F6}-1}\right) - \,(e-2) \\ S = \,(e-1) -\,(e-2) = \boxed{1}$

Try similar problem here .

Observer that each term can be written as $\begin{aligned}\frac1{n!}\ - \frac1{(n+1)!}\end{aligned}$ .

Thus the partial sums are just $\begin{aligned}1 - \frac1{(n+1)!}\end{aligned}$ and so tend to 1.

The Taylor series for $e^x$ is $1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ .

Divide everything by $x$ to get $\dfrac{e^x}{x} = \dfrac{1}{x}+1+\dfrac{x}{2!}+\dfrac{x^2}{3!}+\dfrac{x^3}{4!}+...$ .

Differentiate both sides to get $\dfrac{d}{dx}\dfrac{e^x}{x} = \dfrac{e^x (x-1)}{x^2} = \dfrac{-1}{x^2}+0+\dfrac{1}{2!}+\dfrac{2x}{3!}+\dfrac{3x^2}{4!}+...$ .

Plug in $x = 1$ to obtain $0 = -1 + \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...$ .

Moving the -1 over allows us to obtain the result $1 = \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...$ .

You can solve this without knowing anything about Taylor series or $e$ . Instead, we can solve it by induction.

Let's start by calculating the sum at each step:

1. $\frac {1}{2} = \frac {1}{2}$
2. $\frac {1}{2} + \frac {2}{6} = \frac {5}{6}$
3. $\frac {5}{6} + \frac {3}{24} = \frac {23}{24}$
4. $\frac {23}{24} + \frac {4}{120} = \frac {119}{120}$

We might notice the pattern here. Each step seems to be equal to $\frac {n! - 1}{n!}$ . To prove that this is the pattern, let's use induction.

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Base case

Our base case is the first element in the series: $\frac {1}{2!}$ . Since that's equal to $\frac {2!-1}{2!}$ , our base case matches the pattern we're trying to prove. So we can move on to the induction step.

We could also choose to make the base case $\frac {0}{1!}$ and observe that $\frac {0}{1}$ matches our expected pattern as well. To make the inductive step easier, I will assume that the first term in this series is $\frac {0}{1!}$ , and that $\frac {1}{2!}$ is the second term.

${A}_{1}= \frac {0}{1!}$

${A}_{2}= \frac {1}{2!}$

...

${A}_{n}= \frac {n-1}{n!}$

${A}_{n+1}= \frac {n}{n+1!}$

...

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Induction step

Let's assume that the sum of the first $n$ terms in the series is ${S}_{n} = \frac {n! - 1}{n!}$ . Now we need to show that the sum of the first $n+1$ terms is equal to $\frac {(n+1)! - 1}{(n+1)!}$ .

Let's add ${A}_{n+1}$ to our assumed sum:

${ S }_{ n+1 }={ S }_{ n }+{ A }_{ n+1 }\\ { S }_{ n+1 }={ S }_{ n }+\frac { n }{ (n+1)! } \\ { S }_{ n+1 }=\frac { n!-1 }{ n! } +\frac { n }{ (n+1)! } \\ =\frac { (n!-1)*(n+1) }{ (n+1)! } +\frac { n }{ (n+1)! } \\ =\frac { n!*(n+1)-(n+1)+n }{ (n+1)! } \\ =\frac { (n+1)!-n-1+n }{ (n+1)! } \\ =\frac { (n+1)!-1 }{ (n+1)! }$

Thus the first $n+1$ terms in our series sums to: ${S}_{n+1} = \frac {(n+1)! - 1}{(n+1)!}$ and we have proven our induction step to be true.

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Conclusion

Since our base case and induction step are both true, we conclude that ${S}_{n} = \frac {n!-1}{n!}$ . We can simplify this to ${S}_{n}=1- \frac {1}{n!}$ . So as $n$ goes to infinite, the sum becomes ${S}_{\infty}= 1 - \frac {1}{\infty} = 1 - 0 = 1$ .

Thus our series converges to $\boxed {1}$ .

Another solution (maybe less elegant, but that requires less knowledge) is:

It can easily be shown that:

$\sum\limits_{i=1}^n \frac{i}{(i+1)!} = \frac{1}{2} + \frac{1}{2} \bigg( \frac{2}{3} + \frac{1}{3} \big( \frac{3}{4} + \frac{1}{4} \big(... (\frac{n}{n+1}) \big)\big)\bigg)$

Hence, when $n \to \infty$ , the inner bracket becomes $1$ , hence the one around it... and so on. The sum then becomes:

$\frac{1}{2} + \frac{1}{2} \cdot 1 = 1$

Observe that $S_1=\frac{2!-1}{2!}$ and that generally $S_n=\frac{(n+1)!-1}{(n+1)!}$ (where $S_n$ is the sum of the n first elements)

Prove it using induction if you like to. (I left it out)

Conclude that $\lim_{n\rightarrow \infty}{S_n}=\boxed{1 }$

The basic approch with differentiation, Write expansion of e^x To match the question, divide e^x/ x and take differentiation then substitute x = 1

We will use that:

$e^x=\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$

We will look at the function $f$ below (the idea is to plug $x=1$ ):

$f(x)=\displaystyle\sum_{n=1}^{\infty} (n-1)\frac{x^n}{n!}$ $f(x)=x\times\displaystyle\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} + \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n!}$ $f(x)=xe^x+e^x-1$

Plugging $x=1$ :

$\displaystyle\sum_{n=1}^{\infty}\frac{n-1}{n!}=1 \times e^1 -e^1 +1 = \boxed{1}$

Note: the absolute convergence for all real $x$ of all the series used can be easily proved using comparison with Riemann series.

e^x=1+x+x^2/2!+... therefore (e^x-1)/x=1+x/2!+x^2/3!... suppose x=1 after derivation of equation’s two sides, we got the answer!

Define f(x) = x^2/2! + 2x^3/3! + 3x^4/4! +......... Then f'(x) =x + x^2 + x^3/2! + x^4/3! +....... = X(1 + x + x^2/2! + x^3/3! + ......) =xe^x. Then f(x) = integral of xe^x(dx). Integrating by parts, f(x) = xe^x - e^x + C. Since f(0) =0, C= 1, and f(x) = 1 + xe^x - e^x, and f(1) = 1. Ed Gray

use $e^x$ = $(1+x/1!+x^2/2!+x^3/3!+...$ ) then divide whole equation by $x$ and differentiate wrt to $x$ . now put $x=1$ . the answer comes out to be as 1