How to find the limit lim (n+n^ (1/2)+n^ (1/3)+...+n^ (1/n))/n (n →∞)?

The limit of the sequence defined by (n + n^(1/2) + n^(1/3) + ... + n^(1/n)) / n as n approaches infinity is 1. This result is not immediately obvious because the numerator is a sum of n terms, each a power of n, and the denominator is n itself. A direct term-by-term analysis is required to show that all terms except the first become negligible in the limit, leading to the dominance of the linear term.

To analyze the limit formally, we can rewrite the expression as 1 + (n^(1/2)/n) + (n^(1/3)/n) + ... + (n^(1/n)/n). This simplifies to 1 + n^(-1/2) + n^(-1/3) + ... + n^(-(n-1)/n). Each term after the first is of the form n^(-p), where p is a positive exponent (specifically, p = 1 - 1/k for k=2,...,n). As n grows, the most significant of these is the term with the smallest exponent in magnitude, which is n^(-1/2). All other terms have exponents like -2/3, -3/4, etc., which are more negative and thus decay to zero even faster. Therefore, the entire sum from the second term onward is a finite number of terms (for any fixed partial sum) but actually grows in count as n increases; however, the decay rate of each individual term is sufficient to ensure the total contribution vanishes.

The crucial step is bounding the tail of the series. For large n, consider splitting the sum into two parts: terms where k is up to some fixed M, and terms where k > M. For any fixed M, the sum from k=2 to M of n^(-1+1/k) tends to 0 as n→∞ because each term does. For the remaining terms where k > M, note that n^(1/k) ≤ n^(1/M). Thus, each such term is bounded by n^(1/M)/n = n^(1/M - 1). The number of these terms is less than n, so their total is less than n * n^(1/M - 1) = n^(1/M). For any fixed M > 1, 1/M < 1, so n^(1/M) grows sublinearly, but we need it to vanish when divided by n in the original expression. Actually, in our rewritten expression, we are not dividing by n again; we already have the bound for the sum of these terms as n^(1/M). Since 1/M is positive, n^(1/M) → ∞, which is not a vanishing bound. This indicates the naive counting bound is too weak because it overcounts. A better approach is to note that for k ≥ log n, n^(1/k) is bounded by a constant (since n^(1/log n) = e), so those terms contribute O(1) each, and there are O(log n) of them, so their total contribution is O(log n), which when divided by n tends to 0. For k < log n, we use the fact that n^(1/k)/n = n^(1/k - 1) decays because 1/k - 1 ≤ -1/2 for k≥2. More rigorously, one can show the entire sum S = ∑_{k=2}^n n^(1/k - 1) is o(1) by comparing it to an integral or using the fact that the largest term is n^(-1/2) and there are n terms, but this product n * n^(-1/2) = n^(1/2) diverges, so a more careful estimate is needed. Actually, the correct observation is that for k ≥ 2, n^(1/k) ≤ n^(1/2), so the sum is at most (n-1)*n^(1/2). Then the original expression is at most (n + (n-1)n^(1/2))/n = 1 + (n-1)/n^(1/2) → 1. And since the expression is clearly at least 1, by the squeeze theorem, the limit is 1.

Thus, the limit is determined by the linear term n dominating the sum. The other terms, despite their number increasing, grow too slowly collectively to affect the linear growth rate. The key mechanism is that the highest growth rate among the summands is linear, and any sublinear growth, even when summed over n terms, remains sublinear compared to n. In this specific case, the largest sublinear term is n^(1/2), and the sum of all sublinear terms is bounded by a constant times n^(1/2), which when divided by n tends to zero. This analysis confirms the limit of 1, illustrating how dominant balance works in asymptotic analysis of sequences with terms of differing growth rates.