Is This Convergent Or Divergent Limit Comparison Test

How do I use the limit comparison test to test if 1/n! is convergent?

Note that 1/n! = 1/(1 * 2 * 3 * ... * n) < 1/(1 * 2 * 2 * ... * 2) = 1/2^(n-1) for all n > 2.


So, we next consider the limit
lim(n→∞) (1/n!) / (1/2^(n-1)) = lim(n→∞) 2^(n-1)/n!.

However, this limit equals 0 by the Squeeze Law;
0 < 2^(n-1)/n! < 2^(n-1)/(1 * 2 * 3 * 3 * ... * 3) = (2/3)^(n-2) for n > 3.
Since 2/3 < 1, we have lim(n→∞) (2/3)^(n-2) = 0.

Finally, since lim(n→∞) (1/n!) / (1/2^(n-1)) = 0, and
Σ(n = 0 to ∞) 1/2^(n-1) is convergent (geometric series with |r| = 1/2 < 1),
Σ(n = 0 to ∞) 1/n! must also be convergent by the Limit Comparison Test.

I hope this helps!

Convergence and Divergence using Comparison Test?

Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test). For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.)

1. For all n>1, 1/(nln(n))<2n, and the series 2∑1/n diverges, so by the Comparison Test, the series ∑1/(nln(n)) diverges
2. For all n>2, n/(n^3−8)<2n^2, and the series 2∑1/n^2 converges, so by the Comparison Test, the series ∑n/(n^3−8) converges.
3. For all n>1, arctan(n)/(n^3)<π/(2n^3), and the series π^2∑1/n^3 converges, so by the Comparison Test, the series ∑arctan(n)/(n^3) converges.
4. For all n>1, sin^2(n)/n^2<1/n^2, and the series ∑1/n^2 converges, so by the Comparison Test, the series ∑sin^2(n)/n^2 converges.
5. For all n>1, n^5−n^3<1/n^2, and the series ∑1/n^2 converges, so by the Comparison Test, the series ∑n^5−n^3 converges.
6. For all n>2, 1/(n^2−4)<1/n^2, and the series ∑1/n^2 converges, so by the Comparison Test, the series ∑1/(n^2−4) converges.

I got C, C, I, C, C, I. The computer says its incorrect but I'm not sure which one is wrong.

Nth term test or limit comparison test?

For the sum of (2n)/(3n-1) as n goes to infinity (n starts at 1), the book uses the nth term test for divergence and the limit comes out as 2/3, and thus the series diverges.

For the sum of (10n+1)/[n(n+1)(n+2)] as n goes to infinity (n starts at 1), the book uses the limit comparison test. My question is, why couldn't I have used the nth term test instead? By taking the limit of (10n+1)/[n(n+1)(n+2)], I'd get

(10n+1)/(n^3+3n^2+2n)
divide everything by n^3
and the answer comes out to be 0/1 = 0 so the series converges

But when the book uses the limit comparison test, the limit came out to be 10, not 0.

When do I use one test instead of the other?

Is the series (1+4^n)/(1+3^n) convergent or divergent?

Using direct comparison test: For all n > 0:
(1 + 4^n)/(1 + 3^n) > 4^n/(1 + 3^n) > 4^n/(3^n + 3^n) = (1/2) (4/3)^n.

Since Σ(n=1 to ∞) (1/2)(4/3)^n is a divergent geometric series, the series in question must also diverge.

I hope this helps!

Question about the Limit Comparison Test?

The Limit Comparison test they talk about to you in high school or wherever doesnt really encompass the entire rule. They kind of dumb-down the details for you, no offense. You are correct that there are stipulations for these conditions.

First of all, we take absolute values. It doesnt matter if a_n or b_n are positive or negative. We can take the limit | a_n | / | b_n |. That takes care of the negative limit issue. We are only concerned, now, for limits of 0, positive infinity, or any other finite positive number.

As you already know, if the limit is a finite positive value then both series either converge together, or they diverge together.

What you may not know, and are beginning to suspect, is that the test is a little stronger than that.

If the limit L diverges to infinity, then... only if the limit of |b_n| does NOT equal zero, then the series ∑ a_n also diverges. Makes sense right? The limit would be divergent if the limit of b_n was zero, whether or not ∑ a_n diverged or converged... so we cannot let the limit of b_n to be zero.

Similarly, if the limit a_n / b_n = 0, we know that it could be because the limit of b_n approaches infinity, and it says nothing about ∑ a_n, So if we disallow b_n to approach infinity, but the limit a_n / b_n still approaches zero, then ∑ a_n does converge.

In any other case, the test has failed. You picked a bad series to compare to.

Series Comparison Test, URGENT help?

I would not use sigma signs for any of these comparisons, but that objection aside:

1) False; 1/(n^2 - 1) > 1/n^2.

2) True, via arctan n < pi/2

3) True, using ln n < n^(1/2)

4) True, since 1/(n ln n) < 2/n <==> ln n > 1/2 is true for n = 2, 3, ...

5) True, since n/(n^3 - 8) < 2/n^2 <==> 0 < n^3 - 16 for all n = 3, 4, ...

6) True, since ln n > 1 for all n = 3, 4, ...

I hope this helps!