Lecture
Example N 1
Investigate a series on convergence .
Decision.
Find
Since 2> 1, then, on the basis of Cauchy, the series diverges.
Answer: the series diverges.
Example N 2
Investigate on the convergence series .
Decision.
Since factorials are present in U n , we use the dalamber sign.
Consequently, on the basis of d'Alembert, the series diverges.
Answer: the series diverges.
Example N 3
Investigate on the convergence series .
Decision.
The Cauchy symptom does not work in this case, since . Here we apply the necessary sign of the convergence of series.
Consequently, on the basis of the required characteristic, the series diverges.
Answer: the series diverges.
Example N 4
Investigate on the convergence series
Decision.
Let's make a series of modules of members of our series:
we obtain a sign positive series.
Let's apply the limit sign of comparison. Let's make a row equivalent to a row (*):
series are equivalent. Because - diverges, the series (*) also diverges. Consequently, the original series does not converge absolutely.
We investigate conditional convergence on the basis of Leibniz:
therefore, our series converges conditionally.
Answer: the series converges conditionally.
Example N 5
Find the interval of convergence of the series .
Decision.
Completing the replacement get a row . Find the radius of convergence:
,
Consequently, By performing a reverse replacement, we get . Solve the inequality: , .
Check the convergence of the series on the boundaries of the interval of convergence:
Consequently, the series converges absolutely when in all other points it diverges.
Answer: the series converges absolutely when in all other points it diverges.
Example N 6
Calculate the integral accurate to 0.001.
Decision.
Using Maclaurin's series of functions :
Express the function via functional range:
Since the region of convergence of this series is the interval and spacing , then integrating the obtained series by term, we find:
Received alternating series, therefore, the error in the calculation of the amount does not exceed the modulus of the first rejected member of the series. Find a member of the series, the value of which is less than 0.001:
Therefore, to achieve the required accuracy, it is necessary to drop the member of the series and all the following:
Answer:
Comments
To leave a comment
Ranks
Terms: Ranks