Solution Found!
Find a series Find a series that…a. converges faster than
Chapter 10, Problem 52E(choose chapter or problem)
Find a series Find a series that…
a. converges faster than \(\sum \frac{1}{k^{2}}\) but slower than \(\sum \frac{1}{k^{3}}\).
b. diverges faster than \(\sum \frac{1}{k}\) but slower than \(\sum \frac{1}{\sqrt{k}}\).
c. converges faster than \(\sum \frac{1}{k \ln ^{2} k}\) but slower than \(\sum \frac{1}{k^{2}}\).
Questions & Answers
QUESTION:
Find a series Find a series that…
a. converges faster than \(\sum \frac{1}{k^{2}}\) but slower than \(\sum \frac{1}{k^{3}}\).
b. diverges faster than \(\sum \frac{1}{k}\) but slower than \(\sum \frac{1}{\sqrt{k}}\).
c. converges faster than \(\sum \frac{1}{k \ln ^{2} k}\) but slower than \(\sum \frac{1}{k^{2}}\).
ANSWER:Problem 52E
Find a series Find a series that…
a. converges faster than but slower than
.
b. diverges faster than but slower than
.
c. converges faster than but slower than
.
Solution
Step 1
In this problem we have to find a series in each case that satisfies the given condition.
a. converges faster than but slower than
Let the series to be found be
Thus should satisfy,
Consider
Since 2.5 > 1
By “p-test : diverges if p ≤ 1 and converges if p > 1” we get
converges.
Clearly
Raising to power k, we get
Therefore converges faster than
but slower than
.