Summation of series by definite integrals

To find the summation of the given series as n tends to infinity, we can use definite integrals. Let's break down the process into steps:

Step 1: Rewriting the series
The given series can be rewritten as:
S = lim n→∞ [(1/n)^(1/n) * (2/n)^(1/2n) * (3/n)^(1/3n) * ... * (n/n)^(1/n)]

Step 2: Expressing the series as an integral
We can express the series as an integral by considering the limit of a Riemann sum. Let's divide the interval [0,1] into n equal subintervals of width 1/n. Then, the series can be expressed as:

S = lim n→∞ [(1/n)^(1/n) * (2/n)^(1/2n) * (3/n)^(1/3n) * ... * (n/n)^(1/n)]
= lim n→∞ [(1/n)^(1/n) * (2/n)^(1/2n) * (3/n)^(1/3n) * ... * (n/n)^(1/n)]
= lim n→∞ ∏ (k=1 to n) [(k/n)^(1/kn)]
= lim n→∞ ∏ (k=1 to n) [exp(√(1/kn) * ln(k/n))]

Step 3: Taking the natural logarithm
We can simplify the expression further by taking the natural logarithm of both sides. This allows us to apply the properties of logarithms to simplify the product into a sum:

ln(S) = lim n→∞ ∑ (k=1 to n) [√(1/kn) * ln(k/n)]

Step 4: Converting to a definite integral
To convert the sum into a definite integral, we can approximate each term in the sum as a rectangle with width 1/n and height √(1/kn) * ln(k/n). Then, we can rewrite the sum as an integral:

ln(S) = lim n→∞ ∑ (k=1 to n) [√(1/kn) * ln(k/n)]
≈ ∫ (0 to 1) √(1/x) * ln(x) dx

Step 5: Evaluating the definite integral
Now, we can evaluate the definite integral to find ln(S):

ln(S) = ∫ (0 to 1) √(1/x) * ln(x) dx

This integral can be evaluated using integration techniques or numerical methods.

Step 6: Finding the final summation
Finally, we can find S by taking the exponential of both sides:

S = exp(ln(S))

By evaluating the definite integral in Step 5 and taking the exponential, we can find the summation of the given series as n tends to infinity.