Does 1/ln(x+1) converge or diverge?
I've tried the nth term test, limit comparison test, and integral test. All I get is inconclusive. The other tests I have (geometric series, p-series, telescoping series, alternating series, and root tests) don't apply. What can I use?!
As x-> infinity, 1/ln (x+1) approaches zero. That means that it converges.
Is that the limit you had in mind?
As x-> -1, it converges to zero
As x-> 0, it diverges, since ln 1 = 0
ln (x+1) is undefined for x<-1
No, I wondered if the sum converged, not the sequence. Sorry
No problem! To determine if the series ∑(1/ln(x+1)) converges or diverges, we can use the integral test.
The integral test states that if f(x) is positive, continuous, and decreasing for all x ≥ k, where k is some positive constant, then the series ∑f(n) converges if and only if the corresponding improper integral ∫f(x)dx from k to infinity converges.
In this case, we have f(x) = 1/ln(x+1). Since ln(x+1) is positive, continuous, and decreasing for all x ≥ -1, we can apply the integral test.
To compute the improper integral, we need to determine the limits of integration. We should integrate from -1 to infinity since that is the range where ln(x+1) is defined and f(x) is continuous and decreasing.
Using integral calculus techniques, the indefinite integral of 1/ln(x+1) is ln(ln(x+1)) + C, where C is the constant of integration.
Now, we evaluate the definite integral: ∫[from -1 to infinity] (1/ln(x+1)) dx = [ln(ln(x+1))] evaluated from -1 to infinity.
As x approaches infinity, ln(x+1) increases without bound, so ln(ln(x+1)) also increases without bound. Therefore, the definite integral is divergent.
According to the integral test, since the corresponding improper integral is divergent, the series ∑(1/ln(x+1)) also diverges.
So in conclusion, the series ∑(1/ln(x+1)) diverges.