Problem 10

Question

Let $\left(a_{n}\right)_{n \geq 0}$ and $\left(b_{n}\right)_{n \geq 0}$ be two sequences of complex numbers such that $a_{n}=$ $b_{n}-b_{n+1}, n \geq 0$ Show: The series $\sum_{n=0}^{\infty} a_{n}$ is convergent if and only if the sequence $\left(b_{n}\right)$ is convergent, and then $$ \sum_{n=0}^{\infty} a_{n}=b_{0}-\lim _{n \rightarrow \infty} b_{n+1} $$ Example: $\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}=1 .$

Step-by-Step Solution

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Answer

The series converges if and only if $(b_n)$ is convergent, and $\sum_{n=0}^{\infty} a_{n}=b_{0}-\lim _{n \to \infty} b_{n+1}$.

1Step 1: Understand the Given Relationship

We are given that the sequence $a_n$ is defined as $a_n = b_n - b_{n+1}$. This is the main relation we need to use for the rest of the solution.

2Step 2: Recognize Telescoping Series

Observe that the series $\sum_{n=0}^{\infty} a_n$ becomes a telescoping series due to the relation $a_n = b_n - b_{n+1}$. As you add terms $a_n$, intermediate terms cancel out.

3Step 3: Write the Partial Sums

The partial sum of the series $S_N = \sum_{n=0}^{N} a_n$ becomes $S_N = b_0 - b_{N+1}$. Notice how most terms in the series cancel out.

4Step 4: Condition for Convergence of the Series

The series $\sum_{n=0}^{\infty} a_n$ converges if the remainder of the telescoping series, which is $b_0 - b_{N+1}$, converges to a limit as $N$ tends to infinity.

5Step 5: Link Convergence to Sequence $(b_n)$

For the series to converge, $b_{N+1}$ must converge to a limit as $N$ approaches infinity. This means the sequence $(b_n)$ must be convergent. Therefore, the convergence of $\sum_{n=0}^{\infty} a_n$ is equivalent to the convergence of $\left(b_{n}\right)$.

6Step 6: Final Equality for the Sum

If the sequence $(b_n)$ converges to a limit $L$, then the series sum is $b_0 - L$. Therefore, $\sum_{n=0}^{\infty} a_n = b_0 - \lim_{n \to \infty} b_{n+1}$.

7Step 7: Example Verification

For the example $\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}$, rearrange the term to fit $a_n = b_n - b_{n+1}$, where $b_n = \frac{1}{n+1}$. Notice the sequence $b_n$ converges to 0, confirming the sum $\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)} = 1$.

Key Concepts

Telescoping SeriesSequence ConvergenceComplex Numbers Series

Telescoping Series

A telescoping series is a type of series where many intermediate terms cancel out when summed, simplifying the calculation of the series' sum. This characteristic arises from the specific way terms are constructed, similar to a collapsing domino effect.
In a telescoping series, each term effectively "undoes" the effect of its predecessor, as seen with sequences defined by difference relationships. Consider the given sequence where $a_n = b_n - b_{n+1}$. When the sum $\sum_{n=0}^{\infty}a_n$ is computed, the terms after simplification appear as $b_0 - b_{N+1}$ after canceling intermediate terms.

The essence of telescoping is leveraging these cancellations to find an easily identifiable sum. As $N$ approaches infinity, the value primarily hinges on the behavior of the first and last evident terms $b_0$ and $b_{N+1}$. This simplification is particularly ergonomic when proving convergence as it reduces the sum to a few terms.

Sequence Convergence

Sequence convergence refers to a sequence approaching a specific value as the number of terms increases indefinitely. For a sequence $(b_n)$, convergence means that for any small number, say $\epsilon > 0$, there exists an index $N$ such that for all $n > N$, the difference between $b_n$ and the actual limit $L$ is smaller than $\epsilon$, expressed as $|b_n - L| < \epsilon$.

In our context, to determine the convergence of the series $\sum_{n=0}^{\infty}a_n$, we rely on the convergence of the sequence $(b_{n+1})$, especially as $n$ increasingly approaches infinity. When $b_{N+1}$ approaches a limit $L$, the series convergence is assured by this behavior.

This involves checking that sequence $(b_n)$ settles towards a steady value.
The condition for convergence used in telescoping series is thus whether or not $(b_n)$ converges.

Complex Numbers Series

Complex numbers extend the real line into a plane including imaginary units, represented as $a + bi$ where $a$ and $b$ are real numbers and $i$ is the imaginary unit with $i^2 = -1$. Series involving complex numbers work under the same principles as those with real numbers, but include both real and imaginary components.

For sequences of complex numbers $(b_n)$, as defined in the original exercise, each term generally takes the form of a complex expression $a_n = b_n - b_{n+1}$.

The notion of convergence still applies, where each part (real and imaginary) must converge respectively.
This ensures that the series has a limit not moving outside a neighborhood of a number within the complex plane as $n$ approaches infinity.
When we say that the sequence $(b_n)$ of complex numbers converges, both real parts and imaginary parts must individually converge.

In terms of application, expressing an entire series in such numbers follows the same algebraic manipulations but with respect to complex arithmetic.

Problem 10

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