First slide

Series of natural numbers

Question

Sum of the series $\sum_{r = 1}^{n} \frac{r}{(r + 1)!}$ is

Moderate

A

$1 - \frac{1}{n!}$
B

$1 - \frac{1}{(n + 1)!}$
C

$2 - \frac{1}{(n + 1)!}$
D

none of these

Solution

Let $t_{r}$ denote the rth term of the series, then
$\begin{array}{r} t_{r} = \frac{r}{(r + 1)!} = \frac{r + 1 - 1}{(r + 1)!} = \frac{1}{r!} - \frac{1}{(r + 1)!} \\ \Rightarrow \sum_{r = 1}^{n} t_{r} = \sum_{r = 1}^{n} (\frac{1}{r!} - \frac{1}{(r + 1)!}) = 1 - \frac{1}{(n + 1)!} \end{array}$

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