Table of Contents
Does the integral from 1 to infinity of 1 x converge?
Integral of 1/x is log(x), and when you put in the limits from 1 to infinity, you get log(infinity) – log(1)= infinity -0 = infinity, hence it diverges and gives no particular value. You can think of the integral as a series, sum(1/x) from 1 to infinity which is 1/1+1/2+1/3+1/4+1/5…
Is 1 sqrt x divergent?
int from 1 to infinity of 1/sqrt(x) dx = lim m -> infinity 2sqrt(x) from 1 to infinity = infinity. Hence by the Integral Test sum 1/sqrt(n) diverges.
Does series 1 sqrt and converge?
We also know that ∑∞n=01√n diverges by the integral test and “direct comparison” test.
Does the series 1 sqrt n 1 converge?
The series diverges. ∞∑n=11n is the harmonic series and it diverges. Therefore, by comparison test, ∞∑n=11√n diverges.
Does the improper integral 1 x converge?
if the integrand goes to zero faster than 1x 1 x then the integral will probably converge.
What is the result of 1 infinity?
Infinity is a concept, not a number; therefore, the expression 1/infinity is actually undefined.
How do you find the limit of 1 infinity?
The Solution of 1^Infinity Finding your answer by taking your e function to the power of 0, you get 1. So the limit of your function 2x/3x to the power of x as it goes to infinity is 1.
How do you prove that the integral converges?
Proof. We use the integral test; we have already done p = 1, so assume that p ≠ 1 . ∫ 1 ∞ 1 x p d x = lim D → ∞ x 1 − p 1 − p | 1 D = lim D → ∞ D 1 − p 1 − p − 1 1 − p. If p > 1 then 1 − p < 0 and lim D → ∞ D 1 − p = 0, so the integral converges.
Does the smaller function always converge to infinity?
If the smaller function converges there is no reason to believe that the larger will also converge (after all infinity is larger than a finite number…) and if the larger function diverges there is no reason to believe that the smaller function will also diverge. Let’s work a couple of examples using the comparison test.
What happens if the integrand goes to zero faster than 1?
As noted after the fact in the last section about if the integrand goes to zero faster than 1 x 1 x then the integral will probably converge. Now, we’ve got an exponential in the denominator which is approaching infinity much faster than the x x and so it looks like this integral should probably converge.
How do you make the numerator of an integral larger?
However, we can use the fact that 0 ≤ cos 2 x ≤ 1 0 ≤ cos 2 x ≤ 1 to make the numerator larger ( i.e. we’ll replace the cosine with something we know to be larger, namely 1). So, must also converge. Let’s first take a guess about the convergence of this integral.