Question

Consider the series S=∑ₙ=5[infinity]1/n-n², and its partial sums Sₙ=∑ₙ=5^N1/n-n².

(a) Show that Sₙ=b+bₙ, where b is a constant and bₙ is a rational function of N satisfying b₁₅=1/15 Find the constant b and a formula for bₙ.

Answer 1

To show that Sₙ = b + bₙ, where b is a constant and bₙ is a rational function of N, we need to find the constant b and a formula for bₙ. By evaluating S when n tends to infinity, we find that the constant b is equal to 0. A formula for bₙ can be determined by calculating the partial sums Sₙ for different values of N and observing the pattern.

Step-by-step explanation:

To show that Sₙ = b + bₙ, where b is a constant and bₙ is a rational function of N, we need to find the constant b and a formula for bₙ.

We are given that S = ​∑ₙ=5^∞1/n-n² and Sₙ = ∑ₙ=5^N1/n-n². To find the value of b, we can evaluate S when n tends to infinity.

By using the limit test, we find that S = 1/∞ - ∞² = 0.

Since S is a constant value, we can say that b = 0. Now, let's find a formula for bₙ.

Given b = 0 and b₁₅ = 1/15, we need to find a pattern for bₙ. Using the partial sums Sₙ= ∑ₙ=5^N1/n-n², we can calculate bₙ for different values of N.

Starting with N = 15, we have S₁₅ = ∑ₙ=5^15 1/n-n² = 1/15.

For N = 16, we have S₁₆ = ∑ₙ=5^16 1/n-n² = 1/15 + 1/16.

It can be observed that for each N, Sₙ is equal to the sum of b and a rational function of N. Therefore, we can conclude that Sₙ = b + bₙ, where b = 0 and bₙ is a rational function of N.