Final answer:
Using the sandwich theorem, we establish lower and upper bounds for the series 1/√(n² +1) + ... + 1/√(n² +n), which converge to zero as n approaches infinity. Thus, the limit of the series is zero.
Step-by-step explanation:
To find the limit of the given series 1/√(n² +1) + 1/√(n² +2) + 1/√(n² +3) + ... + 1/√(n² +n) using the sandwich theorem, we need to establish lower and upper bounds that converge to the same limit as n approaches infinity.
Since n² is always less than n²+k for any positive integer k, √(n²) is less than √(n²+k). Thus, 1/√(n²+k) is less than 1/√(n²), giving us a lower bound for each term.
The upper bound for each term can be similarly found by noting that √(n²+k) is less than √(n²+n). By summing these bounds over n terms, we find that the series is bounded between n/√(n²+n) and n/√(n²).
As n goes to infinity, both bounds converge to zero, so by the sandwich theorem, the limit of the series is also zero.