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

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1−1n!

1−1(n+1)!

2−1(n+1)!

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detailed solution

Correct option is B

Let tr denote the rth term of the series, then tr=r(r+1)!=r+1−1(r+1)!=1r!−1(r+1)!⇒ ∑r=1n tr=∑r=1n 1r!−1(r+1)!=1−1(n+1)!

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Sum of the series ∑r=1n r(r+1)! is

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