Solving My Calc 2 Exam#3 (Sequence, Infinite Series & Power Series)

My calculus 2 exam on sequence, infinite series & power series

BIG thanks to vladcmcreates for the timestamps!
00:00 - exam begins, get your pencils out and put your student cards on your desk :P
2:50 - Q1 - Evaluate this infinite series.
3:45 - Q2 - Which of the following series converges absolutely?
8:03 - Q3 - If possible, evaluate the following series.
11:39 - Q4 - Which of the following infinite series diverges by the Test For Divergence?
15:05 - Q5 - Find radius of convergence (R) for the following power series.
18:25 - MISTAKE (but still useful) Q6 - Consider a sequence defined recursively by a1 = 5, an = 8−a for n ≥2. Which of the
following statement about is an true?
22:40 - mistake noticed, followed up by REAL Q6 (Consider a sequence defined recursively by a1 = 5, an = 8−an−1 for n ≥2.)
24:29 - Q7 - Determine the first four nonzero terms of the power series for lnx at a = 2.
Q8 - Integrate the following as a power series and state the radius of convergence.
29:24 - Q8 a)
34:46 - Q8 b)
40:14 - Q9 - Determine if the following series converges or not. Justify your answer.
46:29 - Q11 - Determine if the following series converges or not. Justify your answer.
Q12 - Let an = 2(-3/4)^n.
49:44 - Q12 a) - Does the sequence an converge? If so, to what value?
51:49 - Q12 b) - Does the series an converge? If so, to what value? (HINT: use the ratio test)
53:48 - Q13 - Determine if the following series converges or diverges. Justify your answer.
Q14 - Give an example...
1:00:40 - Q14 a)
1:08:30 - Q14 b)
1:12:28 - Q14 c) and d) (bonus)

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Thank you,
blackpenredpen

[vladimirmlotschek]

TIMESTAMPS [placed right as each question is being written down so you can see what chat was thinking as you solve :) ]:

00:00 - exam begins, get your pencils out and put your student cards on your desk :P
2:50 - Q1 - Evaluate this infinite series.
3:45 - Q2 - Which of the following series converges absolutely?
8:03 - Q3 - If possible, evaluate the following series.
11:39 - Q4 - Which of the following infinite series diverges by the Test For Divergence?
15:05 - Q5 - Find radius of convergence (R) for the following power series.
18:25 - MISTAKE (but still useful) Q6 - Consider a sequence defined recursively by a1 = 5, an = 8−a for n ≥2. Which of the
following statement about is an true?
22:40 - mistake noticed, followed up by REAL Q6 (Consider a sequence defined recursively by a1 = 5, an = 8−an−1 for n ≥2.)
24:29 - Q7 - Determine the first four nonzero terms of the power series for lnx at a = 2.
Q8 - Integrate the following as a power series and state the radius of convergence.
29:24 - Q8 a)
34:46 - Q8 b)
40:14 - Q9 - Determine if the following series converges or not. Justify your answer.
46:29 - Q11 - Determine if the following series converges or not. Justify your answer.
Q12 - Let an = 2(-3/4)^n.
49:44 - Q12 a) - Does the sequence an converge? If so, to what value?
51:49 - Q12 b) - Does the series an converge? If so, to what value? (HINT: use the ratio test)
53:48 - Q13 - Determine if the following series converges or diverges. Justify your answer.
Q14 - Give an example...
1:00:40 - Q14 a)
1:08:30 - Q14 b)
1:12:28 - Q14 c) and d) (bonus)

Watching this stream was such a blast from the past: I just finished Calc 2 in April of 2020. Please let me know if I made any mistakes/if you want me to add anything and I'll fix them right away!
Also, while I'm here, I have to send a huge virtual thank you to you, blackpenredpen: your videos helped me while I took Calc 1 and Calc 2, and even helped me get 100 in my Calc 2 class :D Your videos are always fun to watch, even though I'm no longer taking math courses in university - I'll never stop loving math nor your videos! Thank you SO much!
To current Calc 2 (and Calc 1, if you're hiding) students: work hard and remember to congratulate yourself on every success you make! It's too easy to get lost in the mistakes when you are actually making huge improvements. You can do this!

[treanungkurmal]

3:15 Question 1
5:17 Question 2
9:29 Question 3
13:30 Question 4
16:25 Question 5
20:38 Question 6
25:25 Question 7
30:55 Question 8 (a)
35:14 Question 8 (b)
41:53 Question 9
47:39 Question 11
51:08 Question 12
55:25 Question 13
1:02:20 Question 14 (a)
1:09:55 Question 14 (b)
1:18:43 Question 14 (c & d)

[Peter_]

The Ratio Test is awesome.
I also really like how it ends up being the major test of choice during calculations of Power Series - it's very satisfying how one of the best tests ends up playing a major part like that, while the other tests take a step back and are mostly there as "extra checks" during the last steps for the radius of convergence etc.

[zain]

I absolutely love this guy and this channel!!! Keep it up man !!

[Happy_Abe]

My answer to 14c is the telescoping series from 1 to infinity of a_n where a_n=
((n-1)/n^2)-(n/(n+1)^2)
This can be rewritten as a_n=
(n^2-n-1)/((n^2)(n+1)^2)
This telescoping sum equals:
(n-1)/n^2 at n=1 which is (1-1)/1^2=0
Plus the limit as n approaches infinity of
n/(n+1)^2 which is 0
Thus, 0+0=0 and no value of a_n=0
If anyone finds a mistake please let me know!

[dansnow]

Watched this for review and got a 102 on my exam 3. Thank you blackpenredpen!

[divyakalaria]

Could you create videos for Linear Algebra, because I have the class next semester and I would like to get a head start on it. Plus if I knew about your videos from the start of college I would've watched them for every one of my math classes. Thank You so much for creating great and helpful videos.

[googleit]

Omg. I was watching all his previous video and suddenly jumped to this. HE HAS CHANGED MUCH. I had to match his voice to actually see it was him

[Ironwolf]

Its kinda nice to see that I've been taking calc 2 this current semester, and I'm learning about power series now. I never thought I would be able to understand any of this until I took calc 2 this semester. It's so much fun.

[MrImaddah]

very good calculus after programming class i have just finished :)

[blackpenredpen]

Hi, can someone help with the time stamps? Thank you very much.

[nape]

Hey, BPRP! I recently got an absolutely mysterious equation as a bonus question during a maths test: Solve (x-1)!+1=x^y over the integers. You could do that if you found that interesting.

[cristiano]

Bro, I got a nice question.
It was asked in one of my mathematics exam
Q=>
If ax^2 - bx + c =0 has real and distinct roots lying in
(0, 1) and a, b, c are natural numbers . Find minimum value of a+b+c.
Also state all the assumptions made.
By the way answer is .
a + b +c =11

[seanfraser]

14c: Let a(2n-1) = 1/n and a(2n) = -a(2n-1). That is, {a(n)} = {1, -1, 1/2, -1/2, 1/3, -1/3, ...}

Let S(n) be the nth partial sum. Then for each n, S(2n) = 0 and S(2n-1) = 1/n. So lim S(n) = 0, thus the series converges to 0, even though a(n) =/= 0 for any n.

[Loading]

Cant thank u enough for the time and effort u put for teaching usss

[SartajKhan-jgnz]

Who remembers 14(b) from BPRP challenging Dr. Payem like a year ago? Man the two channels have grown so much!

[TheRandomFool]

As an alternative for #9, you could also use the comparison test by using sin(1/n) is less than or equal to 1/n for large n. Then, the sum of sin^3 (1/n) is less than or equal to the sum of 1/n^3.

[Apollorion]

A potential answer to question 14.c:

a(n)=1/3^n - 1/3^(n+1) - (1/2^n)/3

the first two terms telescope each other out, except for the first 1/3 of a(1). To take that away I subtracted 1/3 of 1, the latter described as the series of 1/2^n .

edit: to verify that a(n) never becomes 0:

which will only be 0 if 2=3 or n+1=0, but since n>0 and 2 will never be 3, this won't be so.
edit 2: I made a mistake above; it should be:
3a(n)=( (3-1)*2^n - 3^n)/6^n = (2*2^n-3^n)/6^n
which will only be 0 if (2/3)^n=1/2 or n=log(1/2)/log(2/3)=( log(1)-log(2) )/(log(2) - log(3)) = log(2)/( log(3)-log(2) ) but can this be solved for a positive integer value?
My calculator app approximates this at a value clearly between 1 and 2, so I guess the answer is: no. . luckily.

[josevidal]

GUYSSS, I found a solution to problem 14c.
Look:
a_n=(-(-1)^(n) ((π)/(2))^(2 (n-1)))/((2n-2)!)

I hope it can be understood, just take the Taylor series of cos(x), modify a bit to make it start at k=1 and evaluate at x=π/2, and THATS IT!

[aurelienhermant]

I have an answer for the really last one.
Take a_n = 1/n^2 and b_n = (1-6/pi^2)^n. That is the most simple I could find