is the sum from k=1 to infinity of 1/(k^2+1) convergent?
i said it was because the integral is convergent
I agree
by the way, every term is smaller than 1/k^2
so the only integral you really have to do is dx/x^2
which is really easy. being -1/x
To determine if the series ∑(k=1 to infinity) 1/(k^2+1) is convergent, you correctly pointed out that we can consider the integral of the function 1/(x^2+1). This approach is based on the integral test, which states that if a function is continuous, positive, and decreasing, then the convergence of the corresponding series can be determined by analyzing the convergence of the integral of the function.
Now, let's explain how to calculate the integral and use it to determine convergence:
1. Obtain the integral:
∫(x=1 to infinity) 1/(x^2+1).
2. Apply basic integration techniques:
The integral can be computed using the substitution method. Let u = x^2 + 1, then du = 2x dx. Therefore, the integral becomes:
(1/2) ∫(u=2 to infinity) 1/u du.
3. Evaluate the integral:
The integral simplifies to:
(1/2) [ln(u)] from 2 to infinity.
Taking the limit as the upper bound approaches infinity, we get:
(1/2) [ln(infinity) - ln(2)].
Since ln(infinity) approaches infinity, this expression diverges.
4. Conclude the convergence of the series:
According to the integral test, if the corresponding integral diverges, then the series also diverges. In this case, since the integral diverges, the series ∑(k=1 to infinity) 1/(k^2+1) also diverges.
Hence, the series is not convergent.