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Find the sum of the series n=1 to [infinity], (3/√(n+1) - 3/√(n+2)) The sum of the series is (Type an exact answer, using radicals as needed.)

User Eric Finn
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Final answer:

The sum of the telescoping series (3/√(n+1) - 3/√(n+2)) from n=1 to infinity is 3/√2, because all terms cancel each other out except for the first term.

Step-by-step explanation:

The student asks to find the sum of the series where the general term is (3/√(n+1) - 3/√(n+2)) from n=1 to infinity. This series is telescoping, meaning each term cancels out part of the following term. To find the sum, we need to write out the first few terms to see the pattern of cancellation:

S = (3/√2 - 3/√3) + (3/√3 - 3/√4) + (3/√4 - 3/√5) + ...

As we can see, -3/√3 from the first term cancels with 3/√3 from the second term, and this pattern continues indefinitely. The terms keep on canceling each other except for the very first term, 3/√2, which never finds a term to cancel with. Therefore, the sum of the series is 3/√2.

User Iamkhush
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