Problem 1
Question
Suppose that the series \(\sum_{k=0}^{\infty} a_{k} x^{k}\) converges at \(x=3 .\) What can you conclude about the convergence or divergence of the following series? (a) \(\sum_{k=0}^{\infty} a_{k} 2^{k}\) (b) \(\sum_{k=0}^{\infty} a_{k}(-2)^{k}\) (c) \(\sum_{k=0}^{\infty} a_{k}(-3)^{k}\) (d) \(\sum_{k=0}^{\infty} a_{k} 4^{k}\)
Step-by-Step Solution
Verified Answer
Using the Ratio Test and comparison, we conclude that:
(a) The series \(\sum_{k=0}^{\infty} a_{k} 2^{k}\) converges.
(b) The series \(\sum_{k=0}^{\infty} a_{k} (-2)^{k}\) converges.
(c) The Ratio Test is inconclusive for the series \(\sum_{k=0}^{\infty} a_{k} (-3)^{k}\).
(d) The series \(\sum_{k=0}^{\infty} a_{k} 4^{k}\) diverges.
1Step 1: Apply the Ratio Test
To apply the Ratio Test, we need to find the limit of the absolute ratio of consecutive terms:
\[L = \lim_{k \to \infty} \left|\frac{a_{k+1}x^{k+1}}{a_kx^k}\right| = |x| \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|\]
If \(L < 1\), the series converges; If \(L > 1\), the series diverges; If \(L = 1\), the test is inconclusive.
2Step 2: Apply the comparison to given series
We're given that \(\sum_{k=0}^{\infty} a_{k} x^{k}\) converges at \(x=3\). So, we can say that:
\[L = 3 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 1\]
Now we can apply the comparison to given series:
(a) \(\sum_{k=0}^{\infty} a_{k} 2^{k}\): The factor is \(x = 2\). Using the Ratio Test:
\[L = 2 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 3 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 1\]
Hence, the series converges.
(b) \(\sum_{k=0}^{\infty} a_{k} (-2)^{k}\): The factor is \(x = -2\). Using the Ratio Test:
\[L = |-2| \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = 2 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 3 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 1\]
Hence, the series converges.
(c) \(\sum_{k=0}^{\infty} a_{k} (-3)^{k}\): The factor is \(x = -3\). Using the Ratio Test:
\[L = |-3| \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = 3 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|\]
Since the series is convergent at \(x=3\), the test is inconclusive at \(x=-3\). We cannot determine if the series converges or diverges.
(d) \(\sum_{k=0}^{\infty} a_{k} 4^{k}\): The factor is \(x = 4\). Using the Ratio Test:
\[L = 4 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| > 3 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 1\]
This implies \(4 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 3 \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|\), which is not true. Therefore, the series diverges.
3Step 3: Conclude your results
- (a) The series \(\sum_{k=0}^{\infty} a_{k} 2^{k}\) converges.
- (b) The series \(\sum_{k=0}^{\infty} a_{k} (-2)^{k}\) converges.
- (c) The Ratio Test is inconclusive for the series \(\sum_{k=0}^{\infty} a_{k} (-3)^{k}\).
- (d) The series \(\sum_{k=0}^{\infty} a_{k} 4^{k}\) diverges.
Key Concepts
Ratio TestConvergence and DivergencePower SeriesLimit Comparison
Ratio Test
The Ratio Test is a powerful tool in determining the convergence of infinite series, especially those with variable terms. This test involves taking the ratio of successive terms in a series. By examining the limit of this ratio as the term number approaches infinity, we can make conclusions about the behavior of the series:
- If the absolute limit is less than 1, the series converges.
- If the limit is greater than 1, the series diverges.
- If the limit equals 1, the Ratio Test cannot determine convergence or divergence.
Convergence and Divergence
A primary focus in the study of series is understanding whether a series converges or diverges. Convergence means that as you sum more terms, the series approaches a specific number. Divergence implies that the series grows indefinitely and doesn't result in a finite value.
- Convergent series have a finite sum as more terms are added.
- Divergent series do not settle to a moving target as more terms are added; they may grow without bound or oscillate.
Power Series
Power series are an intriguing form of series where terms are raised to increasing powers of a variable \(x\). A typical power series might look something like \(\sum_{k=0}^{\infty} a_k x^k\). Understanding these is essential, as they can represent functions, provide approximations, and are relevant in various applications such as physics and engineering.
- Each term of a power series involves multiplying a coefficient \(a_k\) by the variable \(x\) raised to the power \(k\).
- The range of \(x\) values for which a power series converges is known as its "interval of convergence."
- Within this interval, a power series can serve as an infinite polynomial representation of a function.
Limit Comparison
The limit comparison method is another technique used to determine the convergence or divergence of series, often paired with the Ratio Test for more robustness in analysis. It involves comparing the series of interest with a known benchmark series. This works particularly well when directly applying the Ratio Test may be inconclusive or difficult.
- If a known series, \(\sum b_k\), converges, and the limit of the quotient \(a_k/b_k\) as \(k\) approaches infinity is a finite non-zero number, then \(\sum a_k\) converges.
- The same logic applies for divergence; if \(\sum b_k\) diverges and the limit condition holds, then \(\sum a_k\) diverges too.
Other exercises in this chapter
Problem 1
Expand \(f(x)\) in powers of \(x,\) basing your calculations on the geometric series $$\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots$$ $$f(x)=\frac{1}{(1-x)^{2}}$
View solution Problem 1
Find the Taylor polynomial of the function \(f\) for the given values of \(a\) and \(n\) and give the Lagrange form of the remainder. $$f(x)=\sqrt{x} ; \quad a=
View solution Problem 1
Find the Taylor polynomial \(P_{4}\) for the function \(f\) $$f(x)=x-\cos x$$
View solution Problem 1
Determine whether the series converges or diverse. $$\sum \frac{k}{k^{3}+1}$$
View solution