Worksheet 8: Mat 137 Sequences And Series - Full Credit Will
Worksheet 8 Mat 137 Sequences And Seriesnamefull Credit Will Be Given
Worksheet #8 MAT 137 Sequences and Series Name Full credit will be given only where all reasoning and work is provided. Where appropriate, please enclose your final answers in boxes.
1. The Sierpinski Triangle is a fractal that can be constructed as follows: Start with an equilateral triangle and remove an equilateral triangle from the middle. On the next iteration, from each remaining triangle remove a central equilateral triangle as shown below, and continue forever. Assume that the original area of the triangle is A. Construct a sequence that will tell how much area remains after the nth iteration and use this sequence to show that if this process is carried out an infinite number of times the remaining area is 0.
2. (a) Determine the limit of the sequence { 1 / k – 1 / (k + 2) } for k ≥ 1.
(b) Determine the sum of the series ∑k=1∞ (k – 1) / (k + 1).
3. After graduating college, you are offered a job with a starting salary of $50,000 per year. Your contract specifies a guaranteed raise of 4% per year. Over a 40-year career in this job, how much have you made? Note: This is not an infinite series.
4. For what values of z will the geometric series ∑n=0∞ (z + 3)n / 2n converge, and in terms of z, what will it converge to?
5. Series of the form ∑n=0∞ cn / n! are important in mathematics, physics, engineering, computer science, ...
(a) Using the ratio test, show ∑n=0∞ 1 / n! converges.
(b) Using the ratio test, show that ∑n=0∞ cn / n! converges for any constant c ∈ ℝ.
(c) Determine the partial sum s10 = ∑n=09 2n / n! (Write it out but use your calculator.)
(d) Compare your previous answer for s10 to e2 (Use your calculator.)
6. Use the ratio test to show ∑n=1∞ n! / nn is convergent. Hint: You may need the limit definition of e: e := limn→∞ (1 + 1/n)n.
7. If we know ∑n=0∞ an, with an > 0, converges, what can we say about ∑n=0∞ 1 / an? Justify your conclusion.
8. Come up with a counterexample to each statement. That is: each statement is false, provide a specific example of it being false.
(a) ∑ (an – bn) = ∑ an – ∑ bn
(b) ∑ an bn = (∑ an) (∑ bn)
(c) ∑ √ an = √(∑ an)
9. Determine whether the series is absolutely convergent, conditionally convergent, or divergent using an appropriate test:
- (a) ∑n=1∞ n! / n
- (b) ∑n=1∞ (–1)n / n4
- (c) ∑n=3∞ (–1)n ln√ n
- (d) ∑n=1∞ (–1)n 2n / n6
- (e) ∑n=1∞ cos(π n) / n
- (f) 1 + 1 / 2 √2 + 1 / 3 √3 + 1 / 4 √4 + ...
- (g) (ln 2)2 + (ln 3)2 + (ln 4)2 + ...
- (h) ∑k=1∞ √(2k – 1) / 4k2 + 1
Paper For Above instruction
The analysis of sequences and series is fundamental in understanding mathematical structures and convergence behaviors across various disciplines. This paper addresses the series of problems presented in Worksheet 8 for MAT 137, focusing on the properties of geometric and other infinite series, their limits, sums, convergence tests, and real-world applications such as salary growth over time.
Problem 1: The Sierpinski Triangle Area Sequence
The Sierpinski Triangle begins with an equilateral triangle of initial area A. At each iteration, the process involves removing the central smaller equilateral triangle from each remaining triangle, effectively reducing the total area. The pattern of remaining areas can be captured by a geometric sequence. Specifically, after each iteration, the proportion of remaining area is scaled by a factor of 3/4, since a smaller similar triangle with a side length scaled by 1/2 results in an area scaled by 1/4, and three such triangles remain after each iteration, totaling 3/4 of the previous area.
Thus, the sequence describing the remaining area after n iterations is:
Remaining Area = A × (3/4)n
As n approaches infinity, (3/4)n approaches 0 because |3/4|
limn→∞ A × (3/4)n = 0
Therefore, the residual area after infinitely many iterations approaches zero, indicating the fractal's total area shrinks to nothing as the process continues indefinitely.
Problem 2: Limits and Series Sums
(a) Limit of Sequence
The sequence is defined as:
uk = 1 / k – 1 / (k + 2)
As k → ∞, both 1/k and 1/(k + 2) approach 0. The difference thus also approaches 0:
limk→∞ uk = 0
(b) Sum of Series
Consider the series:
∑k=1∞ (k – 1) / (k + 1)
which simplifies to a telescoping series by rewriting numerator and denominator or by partial fraction decomposition. Recognizing the telescoping pattern, the partial sum collapses to the difference of terms at the limits, leading to the sum converging to a finite value, specifically 1.
Problem 3: Salary Growth over 40 Years
The salary each year grows by 4%, forming a geometric series with initial term 50,000 and common ratio 1.04. The total income over 40 years is the sum of this geometric series:
S = 50,000 × (1 – 1.0440) / (1 – 1.04)
Calculating this yields a total income of approximately $3,932,357.52, illustrating how compound interest can significantly increase lifetime earning.
Problem 4: Convergence of Geometric Series in z
The geometric series:
∑n=0∞ (z + 3)n / 2n
has ratio r = (z + 3)/2. For convergence, |r|
|(z + 3)/2|
which simplifies to:
–1
and further to:
–2
resulting in the convergence interval:
–5
Within this interval, the series converges to:
1 / [1 – (z + 3)/2] = 2 / (2 – z – 3) = 2 / (–z – 1)
Problem 5: Series of the form ∑ cn / n!
(a) Convergence of ∑ 1 / n!
Using the Ratio Test, consider:
limn→∞ |(1 / (n + 1)!)/(1 / n!)| = limn→∞ n! / (n + 1)! = limn→∞ 1 / (n + 1) = 0
which confirms convergence.
(b) Convergence for any constant c
Similarly, for ∑ cn / n!, we examine:
limn→∞ |(cn+1 / (n+1)!)/(cn / n!)| = limn→∞ |cn+1/cn| × 1/(n+1)
Provided the ratio |cn+1/cn| is bounded, the limit is zero, ensuring convergence for all real c.
(c) Partial sum s10
The partial sum:
s10 = ∑n=09 2n / n!
Calculations yield an approximate value of 7.388, which can be computed with a calculator.
(d) Approximate e2
The value of e2 is approximately 7.389, so the partial sum s10 closely approximates e2, demonstrating the rapid convergence of the series.
Problem 6: Convergence of ∑ n! / nn
Applying the Ratio Test:
limn→∞ |( (n+1)! / (n+1)n+1 ) / (n! / nn)| = limn→∞ (n+1) nn / (n+1)n+1
which simplifies and approaches 0 due to the properties of e, confirming convergence.
Problem 7: Behavior of reciprocal series
If ∑ an converges with all an > 0, then the series ∑ 1 / an diverges. This is because term-wise reciprocal of a convergent series with positive terms tends to grow without bound, making the reciprocal series divergent.
Problem 8: Counterexamples
(a) Let an = 1n and bn = 1/n. Then, ∑ an divergent while ∑ bn converges, so ∑ (an – bn) does not equal ∑ an – ∑ bn.
(b) Let an = 1n and bn = 1n. Then ∑ an and ∑ bn both diverge, but their product sum is not equal to the product of individual sums.
(c) The series ∑ √ an does not necessarily equal √ (∑ an) unless the series converges absolutely.
Problem 9: Series convergence types
- (a) ∑n=1∞ n! / n: Diverges because factorial growth dominates linear growth.
- (b) ∑n=1∞ (–1)n / n4: Absolutely convergent by the Alternating Series Test and p-series test.
- (c) ∑n=3∞ (–1)n ln√ n: Conditional convergence due to decreasing sequence tending to zero.
- (d) ∑n=1∞ (–1)n 2n / n6: Diverges because 2n grows exponentially.
- (e) ∑n=1∞ cos(π n) / n: Alternating series with term tending to zero, converges conditionally.
- (f) 1 + 1/2 √2 + 1/3 √3 + ...: Series converges as it resembles a p-series with p > 1.
- (g) Sum of (ln n)2: Diverges because logarithmic square grows unboundedly.
- (h) Sum of √(2k – 1) / (4k2 + 1): Converges by comparison to a p-series with p > 1.
Through these analyses, we see the significance of convergence tests such as the ratio, comparison, and p-series tests in classifying series behavior.
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