But converges because it is geometric with . By the DCT, the original series must converge.
(b)
The series has positive terms.
But diverges because it is a -series with . By the DCT, the original series must diverge.
02
The series has positive terms.
Notice that is much greater than for large . So we anticipate that this term will dominate, and we compare the series to .
We seek the limit of this as . Apply L’Hopital’s Rule to the fraction:
Therefore, by continuity:
Since , the LCT says that both series converge or diverge.
Since diverges (), the original series must diverge.
03
(a)
Therefore it is not absolutely converging. Proceed to the AST.
Passes SDT? Yes, .
Decreasing? Yes, denominator is increasing.
Therefore the AST applies and says this converges. So it converges conditionally.
(b)
So this fails the SDT, hence it diverges.
04
(a)
Therefore, by the root test the series converges absolutely.
(b)
Therefore, by the ratio test the series converges absolutely.
(c)
Therefore, by the ratio test the series converges absolutely.
05
(a)
Set . Applicability of the IT:
is is continuous other than at , and the series starts at .
since for all , and for all .
. This is zero at . For , , so the function is decreasing.
Apply the integral test:
This is finite, so the original series converges by the IT.
(b)
For very large , the large powers dwarf the small powers, and the terms look like which equals . So we take this for a comparison series and apply the DCT:
But converges (). So by the DCT, the original series converges.
(c)
For very large , the large powers dwarf the small powers, and the terms look like which equals . So we take this for a comparison series and apply the LCT:
Observe that . Therefore the LCT says that both series converge or both diverge.
We know that converges (). So by the LCT, the original series converges.
06
For very large , we expect the exponentials to dominate, and the series looks like . This will yield a converging geometric series. Anyway, let us choose as the comparison series.
Now divide above and below by the leading power:
By L’Hopital’s Rule, we find that:
Therefore:
Since , the LCT says that both series converge, or both diverge.
Now and this is geometric with . Therefore it converges, and the original series must converge too.
07
(a)
We first check for absolute convergence:
This fails the SDT, so the series diverges!
(b)
Notice that . Check for absolute convergence:
Then:
Since converges (), the DCT says that converges. So the original series converges absolutely and we are done.
08
We use the alternating series test error bound formula. (AKA: “Next Term Bound”)
We seek the smallest such that . What that happens, we will have:
Our formula for :
We cannot easily solve for to provide , so we just start listing out the terms:
We see that is the first term less than 0.005, so and we need the first 5 terms.
09
(a)
Since , the ratio test says that the series diverges.
(b)
Since , the root test says that the series converges.
(c)
Since , the ratio test says that the series converges.