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$The sum to infinity of the series \frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots \dots \dots \dots \dots \dots is equal to:$

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Correct option is A

S=11+11+2+11+2+3+………….tn=11+2+3+……..+n=2n(n+1)=21n−1n+1t1=21−12t2=212−13t3=213−14 and so on tn=21n−1n+1Sn=21−1n+1S∞=2

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The sum to infinity of the series 11+11+2+11+2+3+……………… is equal to:

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$The sum to infinity of the series \frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots \dots \dots \dots \dots \dots is equal to:$