Use the Geometric Series Theory, we know that ∑4n1 convergence, because ∑4n1=∑41⋅41n−1 and 41<1.
Last week, we've learned Comparison Test Theory. So if we can find a series that every element in it is bigger than this one and also converges, then this example must converge, too.
So if n is big enough, then the common ratio will be close to 1/4.
Then we can pick a constant ϵ=0.1 that make ∣an−1an∣<41+0.1, after some calculation we got that n≥15.
So, an+1<(41+0.1)an,wheren≥15,
Then a15+k<(41+0.1)ka15.
We know that ∑(41+0.1)ka15 convergence, base on the Geometric Series Theory.
So ∑a15+k converge.
So ∑ak converge.
1.2. Theorem
Consider n=0∑∞an. And an≥0, n→∞limanan+1=L.
If L<1, then n=0∑∞an converges.
If L>1, then n=0∑∞an diverges.
If L=1, then the ratio test is inconclusive.
Detailed
To find the series (∑n=1∞an,wherean>0) converge or not.
If limn→∞anan+1=L<1,
Then pick a small ϵ so that L+ϵ<1
Find N so that n≥N, anan+1<L+ϵ.
Then aN+k<(L+ϵ)k⋅aN
Since ∑(L+ϵ)k⋅aN converge, ∑aN+k converge.
So ∑ak converge.
If limn→∞anan+1=L>1,
Then pick a small ϵ so that L−ϵ>1
Find N so that n≥N, anan+1>L−ϵ.
Then aN+k>(L−ϵ)k⋅aN
Since ∑(L+ϵ)k⋅aN diverge, ∑aN+k diverge.
So ∑ak diverge.
If limn→∞anan+1=L=1,
Might converge, like ∑n21
Might diverge, like ∑1
1.3. Other Examples
∑nnn!, ∑(n/2)nn!, ∑(n/3)nn!
We know that e=n→∞lim(1+n1)n. With this, we can conclude ∑(n/2)nn! converges, ∑(n/3)nn! diverges.
But what about ∑(n/e)nn!. Let's compare n! with nn
log(n!)=∑k=1nlog(k)≈∫1nlogxdx=xlogx−x]1n
log(n!)=nlogn−n+1
log(n!)≈nlogn−n=log(ennn)
n!≈(en)n
Better approximation is n!≈(en)n2πn, which name is Stirling's approximation.
2. Root Test
2.1. Theorem
an>0,L=n→∞limnan
If L<1,n=1∑∞ converge,
If L>1,n=1∑∞ diverge,
If L=1,n=1∑∞ inconclusive.
3. Integral Test
3.1. Theorem
Suppose f is a continuous, positive, decreasing function on [1,∞), and let an=f(n). Then the series ∑n=1∞an is convergent if and only if the improper integral ∫1∞f(x)dx is convergent. In other words:
If ∫1∞f(x)dx is convergent, then ∑n=1∞an is convergent.
If ∫1∞f(x)dx is divergent, then ∑n=1∞an is divergent.
3.2. Proof of the Integral Test
For the general series ∑an, look at the figures above. The area of the first shaded rectangle is the value of f at the right endpoint of [1, 2], that is, f(2)=a2. So, comparing the areas of the shaded rectangles with the area under y=f(x) from 1 to n, we see that
Notice that this inequality depends on the fact that f is decreasing.
If ∫1nf(x)dx is convergent, then i=2∑nai≤∫1nf(x)dx≤∫1∞f(x)dx since f(x)≥0. Therefore sn=a1+i=2∑nai≤a1+∫1∞f(x)dx
So the sequence {sn} is bounded above.
And {sn} is also an increasing sequence.
This means that ∑an is convergence.
If ∫1nf(x)dx is divergent, then ∫1nf(x)dx→∞asn→∞ because f(x)≥0. But ∫1nf(x)dx≤i=1∑n−1ai=sn−1 and so sn−1→∞. This implies that sn→∞ and ∑an diverges.
3.3. Example
The Harmonic Series ∑n1
The Integral Test need the function is positive and decreasing on [1,∞).
x≥1,f(x)>0
a>b,f(a)<f(b)
Now let's calculate ∫1∞f(x)d(x)
=N→∞lim∫1Nf(x)dx=N→∞lim∫1Nx1dx
=N→∞lim(logN−log1)=N→∞limlogN
=∞
∑np1
Method 1:
With Cauchy Condensation, we know that, ∑n=1∞np1 converges if and only if ∑n=0∞2n(2n)p1 converges.
∑n=0∞2n(2n)p1=∑n=0∞(2n)p−11=∑n=0∞(2p−1)n1
Use the Geometric Series Theory, we got the ratio is 2p−11, so if p<1, it diverges; if p>1, it converges.
Method 2, use The Integral Test:
Let's assume p≠1, which is The Harmonic Series we have proved before.
Also the function is positive and decreasing on [1,∞), so we can use the Integral Test on this.