The value of sum_{r=2}^{infty} frac{1+2+dots+ | vedclass

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(C) The sum of the first $(r-1)$ natural numbers is given by $\frac{(r-1)r}{2}$.Substituting this into the summation, we get $\sum_{r=2}^{\infty} \fra

Vedclass · Accepted Answer

$\frac{e}{2}$

Vedclass · Answer

(C) The sum of the first $(r-1)$ natural numbers is given by $\frac{(r-1)r}{2}$.Substituting this into the summation, we get $\sum_{r=2}^{\infty} \frac{(r-1)r}{2 \cdot r!}$.Since $r! = r(r-1)(r-2)!$, the expression simplifies to $\sum_{r=2}^{\infty} \frac{1}{2(r-2)!}$.Factoring out the constant $\frac{1}{2}$, we have $\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-2)!}$.Let $k = r-2$. As $r$ goes from $2$ to $\infty$, $k$ goes from $0$ to $\infty$.Thus, the expression becomes $\frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{2} e$.

The value of $\sum_{r=2}^{\infty} \frac{1+2+\dots+(r-1)}{r !}$ is:

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